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3 years ago

Which value of m will create a system of parallel lines with no solution?

Mathematics

2 answers:

Nady [450]3 years ago

5 0

Answer:

d.2

Step-by-step explanation:

Pepsi [2]3 years ago

3 0

Rearranging y = 8x -4y = 12 in the y = mx + b form gives y = 2x - 3

y = 2x - 3

y = mx - 6 (already in y = mx + b form)

--------------

Parallel lines have the same slope.

m = 2

Note that the two lines have the different y-intercepts and are not the same line.

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2r+2s=50<br> Please look at the picture

lidiya [134]

2r + 2s = 50

2r - s = 17

-----------subtract

3s = 33

s = 11

2r - s = 17

2r - 11 = 17

2r = 28

r = 14

so r = 14 and s = 11

answer is C. r = 14, s = 11

4 0

3 years ago

$35+0.50()=52 what does this equal

katen-ka-za [31]

Answer:

$(34)$

Step-by-step explanation:

We are looking for what number times .50 makes the equation true.

So, just do backwards equation:

$52 - 35$

$==> 17$

Thus, meaning that what multiplied by 0.50 is equal to 17?

[tex]0.50 * 34[tex]

[tex]==> 17[tex]

6 0

3 years ago

Read 2 more answers

There are five 6-inch sticks and five 10-inch sticks. These sticks can be attached end-to-end to make longer sticks. For example

GalinKa [24]

If you divide all of them together starting inside of the parentheses you will get the wrong answer bc I’m just typing random thing

4 0

3 years ago

Compute the surface area of the portion of the sphere with center the origin and radius 4 that lies inside the cylinder x^2+y^2=

Tom [10]

Answer:

16π

Step-by-step explanation:

Given that:

The sphere of the radius = $x^2 + y^2 +z^2 = 4^2$

$z^2 = 16 -x^2 -y^2$

$z = \sqrt{16-x^2-y^2}$

The partial derivatives of $Z_x = \dfrac{-2x}{2 \sqrt{16-x^2 -y^2}}$

$Z_x = \dfrac{-x}{\sqrt{16-x^2 -y^2}}$

Similarly;

$Z_y = \dfrac{-y}{\sqrt{16-x^2 -y^2}}$

∴

$dS = \sqrt{1 + Z_x^2 +Z_y^2} \ \ . dA$

$=\sqrt{1 + \dfrac{x^2}{16-x^2-y^2} + \dfrac{y^2}{16-x^2-y^2} }\ \ .dA$

$=\sqrt{ \dfrac{16}{16-x^2-y^2} }\ \ .dA$

$=\dfrac{4}{\sqrt{ 16-x^2-y^2} }\ \ .dA$

Now; the region R = x² + y² = 12

Let;

x = rcosθ = x; x varies from 0 to 2π

y = rsinθ = y; y varies from 0 to $\sqrt{12}$

dA = rdrdθ

∴

The surface area $S = \int \limits _R \int \ dS$

$= \int \limits _R\int \ \dfrac{4}{\sqrt{ 16-x^2 -y^2} } \ dA$

$= \int \limits^{2 \pi}_{0} } \int^{\sqrt{12}}_{0} \dfrac{4}{\sqrt{16-r^2}} \ \ rdrd \theta$

$= 2 \pi \int^{\sqrt{12}}_{0} \ \dfrac{4r}{\sqrt{16-r^2}}\ dr$

$= 2 \pi \times 4 \Bigg [ \dfrac{\sqrt{16-r^2}}{\dfrac{1}{2}(-2)} \Bigg]^{\sqrt{12}}_{0}$

$= 8\pi ( - \sqrt{4} + \sqrt{16})$

= 8π ( -2 + 4)

= 8π(2)

= 16π

4 0

3 years ago

On a class trip, there were 45 more girls than boys. the total number of students on the trip was 215. write and solve an equati

Evgesh-ka [11]

(b)+(b+45)=215
b = # of boys
there are 45 more girls than boys and 215 students total.
minus 45 on both sides leaving you with
(b)+(b)=170
combine like terms leaving you with
2b=170
divide both sides by 2
b=85
there were 85 boys
85+45=130
(the 45 for the # of girls more than boys)
there were 130 girls.

3 0

3 years ago