y = x²+1 at x=1 what’s the solution?

Answer:

1. alterative exterior angles.

Step-by-step explanation:

I don't know what the rest of the other options are. So I can't answer the rest.

Answer:

x=-3

Step-by-step explanation:

5(x - 4) + 3x - 9x = 6 - (2x + 5) + 8x

(remove parentheses)

5x-20+3x-9x=6-2x-5+8x

(collect like terms and calculate)

-x-20=1+6x

(move the terms)

-x-6x=1+20

(collect like terms and calculate)

-7x=21

(divide both sides)

x=-3

(your answer :))

Answer:

D

or

It is shifted 2 units up and 2 units right.

Step-by-step explanation:

The graph of  y = sec(x) is shifted 2 units up and 2 units right.

Graph is a mathematical representation of a network and it describes the relationship between lines and points.

The given graph is  y = sec(x)

After transformation it is y = sec(x - 2) + 2

We need to find the transformation applied to y=secx to get  y = sec(x - 2) + 2.

Graph transformation is the process by which an existing graph, or graphed equation, is modified to produce a variation of the proceeding graph.

y=f(x-a) is the same graph shifted 'a' units to the right. If 'a' is negative, then, the graph is shifted to the left.

y = f(x) - a is the same graph, but shifted 'a' units downwards. If 'a' is negative, then the graph will be shifted upwards.

In this case, our main function is y=sec(x). And the function y= sec(x-2)+2 is shifted two units to the right and 2 units upwards.

Hence, the graph of  y = sec(x) is shifted 2 units up and 2 units right.

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Answer:

The value is  [tex]p-value = 0.094963[/tex]

Step-by-step explanation:

From the question we are told that

   The sample size is  n  = 500

   The number of rejected parts is  k =  10

    The null hypothesis is [tex]H_o : p = 0.03[/tex]

    The alternative hypothesis is  [tex]H_1 : p < 0.03[/tex]

Generally the sample proportion is mathematically represented as

           [tex]\^ p = \frac{10}{500}[/tex]

=>       [tex]\^ p = 0.02[/tex]

Generally the test statistics is mathematically represented as

           [tex]z = \frac{\^ p - p }{ \sqrt{\frac{p (1-p)}{n} } }[/tex]

=>       [tex]z = \frac{0.02 - 0.03 }{ \sqrt{\frac{ 0.03 (1-0.03)}{500} } }[/tex]

=>       [tex]z = -1.31080[/tex]

From the z table the probability of  (Z <   -1.31080) is  

      [tex]p-value = P(Z < -1.31080) = 0.094963[/tex]

84 I'm sure it's that I did it before

Answer:

555052

Step-by-step explanation:

Answer: it would be 3000 that’s what it is

Step-by-step explanation:

Answer:

31.x>2

32.x<-2

33.x<5

34.x>0

Step-by-step explanation:

Answer:

9900 watts-hours is how much it consumes in a period of 5 days and 12 hours.............to check this: multiply 1800 by 5 and it equals 9000 then you divide 1800 by 2 because 12hours is a half day and it is 900 after all that you add 9000+900 and it equals 9900

Mark me as brainliest if u found the answer helpful

Answer: well just do the math 1800 times 24 is 43200 times 5 is 216000

and 1800 times 6 is 10800 plus 21600 is.......226800

Step-by-step explanation:

do the math!!!!!

Answer:

Part A: -8

Part B: -4

Step-by-step explanation:

[tex]5-13=-8[/tex]

[tex]-8+4=-4[/tex]

Answer:

Step-by-step explanation:

[tex]\frac{f(b)-f(a)}{b-a}[/tex] This is the formula for the average rate of change

over the interval (a, b)

a and b are x values

f(a) ad f(b) think of them as y values

Start calculating f(a) and f(b) a=1 b=6

replace all x's in formula with a, (1)

f(1)=1³-9(1) = 1-9= -8    f(1) = -8

Now do : b (6)

f(6) = 6³ - 9 (6) = 216-54=162 f(6) = 162

now plug everything into your equation

[tex]\frac{f(6)-f(1)}{6-1} =\frac{162- (-8)}{5} =\frac{170}{5} =34[/tex] average rate of change.

notice the 162 - (-8) In math there is no subtraction, and you would change it to + and when you do that you much change the sign of the next value making -8 a +8.

Hope this helps. Good luck!

Answer: false

Step-by-step explanation:

Answer:

No. √14 + 6.2 is more than 3π - 8.2

Step-by-step explanation:

√14 + 6.2

√20.2

≈ 4.494

π ≈ 3.14159

3(3.14159) - 8.2

9.42577 - 8.2

1.22477

Therefore, the answer is no.

Answer:

Is there supposed to be an image?

Step-by-step explanation:

Answer:

b

Step-by-step explanation:

Answer:

12 c+10

Step-by-step explanation:

becus

Answer:

20

Step-by-step explanation:

Answer:

-10, 0, 10, 20, 30, etc.

Step-by-step explanation:

There are multiple elevations greater than -20

If you add +10 you will get closer to the positives. Any positive is greater than -20, EVEN 0!

Answer:

32

Step-by-step explanation:

if you add 4- + 36 it's exactly the same as saying 4 - 36 = 32

pls mark brainlest for real :,,(

Answer:

you clean!?

Step-by-step explanation:

I just lick it

Answer:

use toilet paper

Step-by-step explanation:

first you clean yourself then you clean again until very clean lol

Answer: wow that's crazy cause I need help with the same answer

Step-by-step explanation:

Answer: Points are not following a straight-line pattern

Step-by-step explanation:

Histograms are used to show the distribution of a dataset.

If the histogram is skewed, then it means that the points do not follow a straight-line pattern, on the normal quantile plot.

From the question, we understand that:

The histogram is skewed

This means that:

The data points of the histogram are either concentrated on the right or on the left-hand side of the histogram.

The above highlight mean that:

The points on the histogram are unequal.

Hence, you would see that the points do not follow a straight line pattern on the normal quantile plot.

Read more about normal quantile plot at:

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Answer:

[tex]T = (20,25)[/tex]

Step-by-step explanation:

Given

[tex]S = (-5,0)[/tex]

[tex]M = (10,15)[/tex]

[tex]Ratio = 3:2[/tex]

Required

Determine the coordinate of T

Line division by ratio is calculated using:

[tex]M(x,y) = (\frac{mx_2 + nx_1}{m+n},\frac{my_2 + ny_1}{m+n})[/tex]

Where:

[tex](x_1,y_1) = (-5,0)[/tex]

[tex](x,y) = (10,15)[/tex]

[tex]m:n = 3:2[/tex]

So, we have:

[tex](10,15) = (\frac{3 * x_2 + 2 * -5}{3+2},\frac{3 * y_2 + 2 * 0}{3+2})[/tex]

[tex](10,15) = (\frac{3x_2 -10}{5},\frac{3y_2}{5})[/tex]

Multiply through by 5

[tex]5 * (10,15) = (\frac{3x_2 -10}{5},\frac{3y_2}{5}) * 5[/tex]

[tex](50,75) = (3x_2 -10,3y_2)[/tex]

By comparison:

[tex]3x_2 - 10 = 50[/tex]

[tex]3y_2 = 75[/tex]

[tex]3x_2 - 10 = 50[/tex]

[tex]3x_2 = 50 + 10[/tex]

[tex]3x_2 = 60[/tex]  --- Divide through by 3

[tex]x_2 = 20[/tex]

[tex]3y_2 = 75[/tex]

[tex]y_2 =75/3[/tex] --- Divide through by 3

[tex]y_2 =25[/tex]

Hence:

The coordinates of T is

[tex]T = (20,25)[/tex]

9514 1404 393

Answer:

  x = -0.25+√2.4375i or -0.25-√2.4375i

Step-by-step explanation:

The quadratic formula works for this. The quadratic equation ...

  ax² +bx +c = 0

has solutions given by the quadratic formula:

  [tex]x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}[/tex]

Your equation ...

  2x² +x +5 = 0

has coefficients a=2, b=1, c=5, so the quadratic formula tells you the solutions are ...

  [tex]x=\dfrac{-1\pm\sqrt{1^2-4\cdot2\cdot5}}{2\cdot2}=\dfrac{-1\pm\sqrt{-39}}{4}\\\\x=\left\{-\dfrac{1}{4}+\dfrac{\sqrt{39}}{4}i, -\dfrac{1}{4}-\dfrac{\sqrt{39}}{4}i\right\}[/tex]

The decimal equivalents are ...

  x = -0.25+√2.4375i   or   -0.25-√2.4375i . . . . . . the 'i' is outside the radical

Answer:

5

Step-by-step explanation:

7*5=35

12*5=60

20*5=100

Answer:

j=v-k/a

Step-by-step explanation:

It's like order of operations backwards. So first, you would get rid of the k, by subtracting it from both sides. Then, to isolate j you would divide both sides by a.

Answer:

The sign of the leading coefficient for the polynomial equation of the graph is NEGATIVE

Hope this helps!

Answer: 1 turning point

Step-by-step explanation:

Answer:

for r= 70 cm

the rate is 0.008cm/s

for r=95cm

the rate= 0.0044cm/s

Step-by-step explanation:

kindly find attached a rough sketch of the circle and a detailed solution of the problem.

Given data

rate dv/dt=500cm^3

r=70cm

r=95cm

Answer:

The area of this circle is [tex](\frac{\pi}{2} )[/tex]  the area of the square.

For the uniform electric field normal to the surface, the flux through the surface is electric field multiplied by the area of this surface.

Therefore, Φsquare is [tex](\frac{2}{\pi} )[/tex] ϕcircle

Step-by-step explanation:

Area of the circle is given by;

[tex]A_c = \frac{\pi d^2}{4}[/tex]

Area of the square is given by;

[tex]A_s = L^2[/tex]

relationship between the edge length of the square, d, and length of its side, L,

[tex]d = \sqrt{L^2 + L^2} \\\\d = \sqrt{2L^2}[/tex]

But area of the square , [tex]A_s = L^2[/tex]

[tex]d = \sqrt{2A_s}[/tex]

Then, the area of the square in terms of the edge length is given by;

[tex]A_s = \frac{d^2}{2}[/tex]

Area of the circle in terms of area of the square is given by;

[tex]A_c = \frac{\pi d^2}{4} = \frac{\pi}{2}(\frac{d^2}{2} )\\\\But \ A_s = \frac{d^2}{2} \\\\A_c = \frac{\pi}{2}(\frac{d^2}{2} )\\\\A_c = \frac{\pi}{2}(A_s )[/tex]

For the uniform electric field normal to the surface, the flux through the surface is electric field multiplied by the area of this surface.

Ф = E.A

Flux through the surface of the circle is given by;

[tex]\phi _{circle} = E.(\frac{\pi d^2}{4})[/tex]

Flux through the surface of the square is given by;

[tex]\phi _{square} = E.(\frac{d^2}{2} )\\\\\phi _{square} =E.(\frac{d^2}{2} ).(\frac{\pi}{2} ).(\frac{2}{\pi} )\\\\\phi _{square} =E.(\frac{\pi d^2}{4} ).(\frac{2}{\pi} )\\\\\phi _{square} =(\phi _{circle}).(\frac{2}{\pi} )[/tex]

Therefore, Φsquare is [tex](\frac{2}{\pi} )[/tex] ϕcircle

If the circle has the same diameter as the edge length of the square, then the area of this circle is [tex]\rm \dfrac{\pi }{2}[/tex] the area of the square

The uniform electric field is normal to the surface, the flux through the surface is the electric field multiplied by the area of this surface.

Φsquare is [tex]\dfrac{2}{\pi}[/tex]  ϕcircle

If the circle has the same diameter as the edge length of the square, then the area of this circle is ___________the area of the square

For the uniform electric field normal to the surface, the flux through the surface is____________the area of this surface.

Therefore, Φsquare is ________ ϕcircle .

1. If the circle has the same diameter as the edge length of the square, then the area of this circle is ___________the area of the square.

The area of the circle is;

[tex]\rm Area \ of \ circle = \dfrac{\pi d^2}{4}[/tex]

The area of the square is;

[tex]\rm Area \ of \ square = a^2[/tex]

The relationship between the edge length of the square, d, and length of its side a is;

[tex]\rm d = \sqrt{a^2+a^2}\\\\d = \sqrt{2a^2}\\\\d = \sqrt{2} a[/tex]
The area of the circle in terms of the area of the square is;

[tex]\rm Area \ of \ circle = \dfrac{\pi }{2} \ Area \ of \ square[/tex]

If the circle has the same diameter as the edge length of the square, then the area of this circle is [tex]\rm \dfrac{\pi }{2}[/tex] the area of the square.

2. The uniform electric field is normal to the surface, the flux through the surface is the electric field multiplied by the area of this surface.

Ф = E.A

3. Flux through the surface of the circle is given by;

[tex]= \rm E\dfrac{\pi d^2}{4}\\\\=\dfrac{2}{\pi }[/tex]

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