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2 years ago

9

HELP ASAP TIMED!!!!!!!!!!!!!!!

1 answer:

jenyasd209 [6]2 years ago

3 0

Step-by-step explanation:

1 ans 35/37

2 ans 0.95

3 ans 12/37

4 ans 0.32

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HELP ILL GIVE BRAINLISTTT!

lions [1.4K]

Answer:

You can substitute

4

as

x

in the first equation because

x

and

4

are equal.

2

(

4

)

+

y

=

29

8

+

y

=

29

y

=

21

Step-by-step explanation:

8 0

2 years ago

What is the slope of the line that passes through (-2, 7) and (4, 9)? user: which of the following represents 3x - 5y 10 = 0 wri

patriot [66]

The slope is (9-7)(4-(-2)) = 2/6 = 1/3;

<span> user: which of the following represents 3x - 5y 10 = 0 written in slope-intercept form???
</span>
The correct anwer for the last question is b).

3 0

3 years ago

Read 2 more answers

How do you do systems of equations ?

Arada [10]

There are 2 methods you are taught in Middle School.

Elimination and Substitution.

8 0

3 years ago

The equation of a circle in general form is x2+y2+22x+14y−55=0 . What is the equation of the circle in standard form?

ra1l [238]

$x^2+y^2+22x+14y-55=0 \\\\x^2+22x+y^2+14y=55\\\\(x^2+22x+121)-121+(y^2+14y+49)-49=55\\\\(x+11)^2+(y+7)^2-121-49=55\\\\
(x+11)^2+(y+7)^2-170=55\\\\(x+11)^2+(y+7)^2=55+170\\\\\boxed{(x+11)^2+(y+7)^2=225}$

Answer B.

4 0

2 years ago

Read 2 more answers

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