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3 years ago

7

Find the slope of the line graphed below

2 answers:

Rainbow [258]3 years ago

8 0

Using the form rise/run your graph has a slope of 1/2 with a y-intercept of 1

irina1246 [14]3 years ago

7 0

Answer:

Up one over two = m<rise/run = 1/2

Step-by-step explanation:

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Help me please. Not sure how to solve this. Can someone show me how to do it step by step. -2/5x-9<9/10

Elena L [17]

Leave the x alone first. -2/5x<9/10+9. then -2/5×<99/10. multiply both sides by 5 you'll get -99/2 < 2x. then divide both sides by 2 you'll get -99/4 < ×

7 0

3 years ago

What is 7/8 times a number. the product is less than 7/8 tho

zzz [600]

Answer:

7/8 x < 7/8

Step-by-step explanation:

4 0

2 years ago

Which of the sets of ordered pairs represents a function?

sergey [27]

Answer:

Both A and B

Step-by-step explanation:

A set of ordered pairs is not a function if there is a repeating output (Y value.)

Y values of A= 5, 2, -2, -5.

Y values of B= 2, -2, 4, -3.

There are no repeating outputs.

4 0

2 years ago

In two or three sentences explain how you would solve for the real solutions of the following equation: please help asap!!

g100num [7]

the real solutions for the equation $x^{3}=-64$ are -

$x=4,2+-2\sqrt{3i}$

Step-by-step explanation:

$x^{3}$ = $- 64$

$x^{3} +64$ = 0

We can write 64 as $4^{3}$

$x^{3}$ + $4^{3}$ = 0

using the identity ( $x^{3}+y^{3} = (x+y)(x^{2} -xy+y^{2} )$ )

we get,

= $(x+4) (x^{2} -x*4+4^{2} )$

= $(x+4)(x^{2} -4x+16)$ ....................(1)

solving the quadratic equation ,

$x^{2} -4x+16$ =0

solutions of this quadratic equation can be obtained by

$x=-b +- \sqrt{b^{2}-4ac } /2a$

let use y for factors

$x=-(-4x)+-\sqrt{(-4x^{2} )-4*x^{2} *16} / 2*x^{2}$

$x=4x+-\sqrt{16x^{2} -64x^{2} } /2x^{2}$

$x=4+-\sqrt{16-64}/2$

$x=4+-4\sqrt{3i} /2$

<u /> $x=2+-2\sqrt{3i}$ ..................(2)

from the equation 1 we have,

$x-4=0$

which gives solution $x=4$

and from equation 2 we got $x=2+-2\sqrt{3i}$

so the real solutions for the equation $x^{3}=-64$ are -

$x=4,2+-2\sqrt{3i}$

3 0

3 years ago

In each of the following ,make y the subject and hence find the value of y when a=2, b=3 ,and c=4.

zvonat [6]

Part i

$\dfrac{2a-7}{6y} = \dfrac{3b-5}{2c}$

$y = \dfrac{ 2c(2a-7)}{6(3b-5)} = \dfrac{c(2a-7)}{9b-15}$

Substituting,

$y = \dfrac{4(2(2)-7)}{9(3)-15} = \dfrac{ -12}{12} = -1$

Answer: -1

Part ii

I'm not sure that one's typed in correctly but I'll solve it as written.

$3-34y + 2a = \dfrac{3b-5}{2c + 1}$

$34y = 3+2a-\dfrac{3b-5}{2c + 1}$

$y = \frac 1 {34}\left(3+2a-\dfrac{3b-5}{2c + 1} \right)$

We're not asked to simplify it so I wont. Substituting,

$y = \frac 1 {34}\left(3+2(2)-\dfrac{3(3)-5}{2(4) + 1} \right) = \frac 1 {34}(7-4/9) = \dfrac{59}{306}$

Answer: 59/306

3 0

3 years ago