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2 years ago

12

7th grade math help pls it’s about slope and stuff

1 answer:

Musya8 [376]2 years ago

8 0

Answer:

m=-2

entire equation in slop int. form is: 7=-2x+5

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Please help, if you do not know do not answer please!

Paul [167]

Answer:

1,3 are functions

2,4 are not

6 0

2 years ago

Read 2 more answers

A bag contains 30 red and blue marbles. If 19 are blue, what fraction of the marbles are blue?

Musya8 [376]

Answer:

19%

Step-by-step explanation:

19/100. i don't think it is that simple, but i did the best i could

5 0

2 years ago

Carlos drew a plan for his garden on a coordinate plane. Rose bushes are located at A(–5, 4), B(3, 4), and C(3, –5)

BartSMP [9]

Given:

A(-5,4)

B(3,4)

C(3,-5)

So point D is:

so point D is (-5,-5)

For AB is

Distance between two point is:

$\begin{gathered} (x_1,y_1)and(x_2,y_2) \\ D=\sqrt[]{(x_2-x_1)^2+(y_2-y_1)^2} \end{gathered}$

so distance between A(-5,4) and B(3,4) is:

$\begin{gathered} D=\sqrt[]{(3-(-5))^2+(4-4)^2} \\ =\sqrt[]{(8)^2+0^2} \\ =8 \end{gathered}$

So AB is 8 unit apart.

For B(3,4) and C(3,-5).

$\begin{gathered} D=\sqrt[]{(3-3)^2+(-5-4)^2} \\ =\sqrt[]{0^2+(-9)^2} \\ =9 \end{gathered}$

So BC is 9 unit apart.

For fourth bush point is (-5,-5) it left of point C(3,-5) is:

$\begin{gathered} D=\sqrt[]{(3-(-5))^2+(-5-(-5))^2} \\ =\sqrt[]{(8)^2+0^2} \\ =8 \end{gathered}$

so fourth bush is 8 unit left of C.

For fourth bush(-5,-5) below to point A(-5,4)

$\begin{gathered} D=\sqrt[]{(-5-(-5))^2+(4-(-5))^2} \\ =\sqrt[]{0^2+9^2} \\ =9 \end{gathered}$

so fourth bush 9 units below of A.

8 0

10 months ago

What is the slope of a line that passes through (-12,15) and (0,4)

pav-90 [236]

Answer:

D. -11/12

Step-by-step explanation:

7 0

3 years ago

D<br> Evaluate<br> arcsin<br> (6)]<br> at x = 4.<br> dx

sineoko [7]

Answer:

$\frac{1}{2\sqrt{5} }$

Step-by-step explanation:

Let, $\text{sin}^{-1}(\frac{x}{6})$ = y

sin(y) = $\frac{x}{6}$

$\frac{d}{dx}\text{sin(y)}=\frac{d}{dx}(\frac{x}{6})$

$\frac{d}{dx}\text{sin(y)}=\frac{1}{6}$

$\frac{d}{dx}\text{sin(y)}=\text{cos}(y)\frac{dy}{dx}$ ---------(1)

$\frac{1}{6}=\text{cos}(y)\frac{dy}{dx}$

$\frac{dy}{dx}=\frac{1}{6\text{cos(y)}}$

cos(y) = $\sqrt{1-\text{sin}^{2}(y) }$

= $\sqrt{1-(\frac{x}{6})^2}$

= $\sqrt{1-(\frac{x^2}{36})}$

Therefore, from equation (1),

$\frac{dy}{dx}=\frac{1}{6\sqrt{1-\frac{x^2}{36}}}$

Or $\frac{d}{dx}[\text{sin}^{-1}(\frac{x}{6})]=\frac{1}{6\sqrt{1-\frac{x^2}{36}}}$

At x = 4,

$\frac{d}{dx}[\text{sin}^{-1}(\frac{4}{6})]=\frac{1}{6\sqrt{1-\frac{4^2}{36}}}$

$\frac{d}{dx}[\text{sin}^{-1}(\frac{2}{3})]=\frac{1}{6\sqrt{1-\frac{16}{36}}}$

$=\frac{1}{6\sqrt{\frac{36-16}{36}}}$

$=\frac{1}{6\sqrt{\frac{20}{36} }}$

$=\frac{1}{\sqrt{20}}$

$=\frac{1}{2\sqrt{5}}$

4 0

2 years ago