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3 years ago

12

What is the y-intercept? (20,0) 0 (0, 18) (18,0) O (0,20)

1 answer:

Y_Kistochka [10]3 years ago

4 0

The answer is (0,18)

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A 7th-grade class has 14 girls and 10 boys. An 8th-grade class has 11 girls and 13 boys. What is the probability of picking a st

melomori [17]

Answer:

14>10 11<13

Step-by-step explanation:

4 0

3 years ago

SOMEONE HELP ME PLSSSSSSS

mr Goodwill [35]

Answer:

y = 8x

Step-by-step explanation:

hope this is what u need if not comment and I'll change it

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3 years ago

The number of views on an interesting video after it's uploaded is represented by the following table:

defon

Answer:

The model for V(t) ,the number of views, t ,months after it's uploaded, best fits the data is $,V(t)=50(6.25)^t$

Step-by-step explanation:

Time (months) Views

0 50

2 313

4 1950

6 12,210

8 76,300

10 476,800

$(x_1,y_1)=(0,50)\\(x_2,y_2)=(2,313)\\(x_3,y_3)=(4,1950)$

$Slope = \frac{y_2-y_1}{x_2-x_1}$

$Slope = \frac{313-50}{2-0}$

Slope =131.5

$Slope = \frac{y_3-y_2}{x_3-x_2}$

$Slope = \frac{1950-313}{4-2}$

Slope =818.5

Since the slope is not same . So, it is not a linear relationship

$\frac{y_2}{y_1}=\frac{313}{50}=6.25\\\frac{y_3}{y_2}=\frac{1950}{313}=6.25$

Since the ratio between the consecutive outputs is same so, it is exponential relationship.

Formula : $y=ab^t$

Where a is the initial value = 50

b = rate of change = 6.25

The model for V(t) ,the number of views, t ,months after it's uploaded, best fits the data

So $,V(t)=50(6.25)^t$

5 0

3 years ago

Match the identities to their values taking these conditions into consideration sinx=sqrt2 /2 cosy=-1/2 angle x is in the first

BaLLatris [955]

Answer:

$\cos(x+y)$ goes with $-\frac{\sqrt{6}+\sqrt{2}}{4}$

$\sin(x+y)$ goes with $\frac{\sqrt{6}-\sqrt{2}}{4}$

$\tan(x+y)$ goes with $\sqrt{3}-2$

Step-by-step explanation:

$\cos(x+y)$

$\cos(x)\cos(y)-\sin(x)\sin(y)$ by the addition identity for cosine.

We are given:

$\sin(x)=\frac{\sqrt{2}}{2}$ which if we look at the unit circle we should see

$\cos(x)=\frac{\sqrt{2}}{2}$ .

We are also given:

$\cos(y)=\frac{-1}{2}$ which if we look the unit circle we should see

$\sin(y)=\frac{\sqrt{3}}{2}$ .

Apply both of these given to:

$\cos(x+y)$

$\cos(x)\cos(y)-\sin(x)\sin(y)$ by the addition identity for cosine.

$\frac{\sqrt{2}}{2}\frac{-1}{2}-\frac{\sqrt{2}}{2}\frac{\sqrt{3}}{2}$

$\frac{-\sqrt{2}}{4}-\frac{\sqrt{6}}{4}$

$\frac{-\sqrt{2}-\sqrt{6}}{4}$

$-\frac{\sqrt{6}+\sqrt{2}}{4}$

Apply both of the givens to:

$\sin(x+y)$

$\sin(x)\cos(y)+\sin(y)\cos(x)$ by addition identity for sine.

$\frac{\sqrt{2}}{2}\frac{-1}{2}+\frac{\sqrt{3}}{2}\frac{\sqrt{2}}{2}$

$\frac{-\sqrt{2}+\sqrt{6}}{4}$

$\frac{\sqrt{6}-\sqrt{2}}{4}$

Now I'm going to apply what 2 things we got previously to:

$\tan(x+y)$

$\frac{\sin(x+y)}{\cos(x+y)}$ by quotient identity for tangent

$\frac{\sqrt{6}-\sqrt{2}}{-(\sqrt{6}+\sqrt{2})}$

$-\frac{\sqrt{6}-\sqrt{2}}{\sqrt{6}+\sqrt{2}}$

Multiply top and bottom by bottom's conjugate.

When you multiply conjugates you just have to multiply first and last.

That is if you have something like (a-b)(a+b) then this is equal to a^2-b^2.

$-\frac{\sqrt{6}-\sqrt{2}}{\sqrt{6}+\sqrt{2}} \cdot \frac{\sqrt{6}-\sqrt{2}}{\sqrt{6}-\sqrt{2}}$

$-\frac{6-\sqrt{2}\sqrt{6}-\sqrt{2}\sqrt{6}+2}{6-2}$

$-\frac{8-2\sqrt{12}}{4}$

There is a perfect square in 12, 4.

$-\frac{8-2\sqrt{4}\sqrt{3}}{4}$

$-\frac{8-2(2)\sqrt{3}}{4}$

$-\frac{8-4\sqrt{3}}{4}$

Divide top and bottom by 4 to reduce fraction:

$-\frac{2-\sqrt{3}}{1}$

$-(2-\sqrt{3})$

Distribute:

$\sqrt{3}-2$

6 0

3 years ago

Three of nine softball players play outfield.what fraction of the players play outfield? Write the answer as a reduced fraction

crimeas [40]

3/9 is the original fraction that play outfield
The reduced fraction is 1/3
3/9 = 1/3

5 0

3 years ago