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3 years ago

Why is 1+(-5) equal to -4

Mathematics

1 answer:

erica [24]3 years ago

5 0

1 + (-5) becomes 1 - 5 because you are adding a negative number.

1 - 5 is -4

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Help plz. It’s number 6. Thanks

julsineya [31]

A = 1/2 h (a+b)
A / 1/2 h = a + b
A / 1/2 h - b = a

a = A / 1/2 h - b

Hope this helps

7 0

2 years ago

Can someone help me please? I don't nderstand this

musickatia [10]

Answer:

perpendicular(p)=5

base(b)=12

hypotenuese(h)=?

using pytha goras theorem,

h^2=P^2+b^2

h^2=(5)^2+(12)^2

h^2=25+144

h^2=169

h^2=13^2

h=13.

6 0

2 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

2 years ago

Is 7.5 greater or less than 48.75?

fomenos

Answer:

Less than

Step-by-step explanation:

7.5 < 48.75

48.75 is a bigger number than 7.5

Therefore, 7.5 is less than it.

7 0

3 years ago

Read 2 more answers

Solve for B and A please help ASAP

Mama L [17]

Answer:

∠A = ∠B = 80°

Step-by-step explanation:

The angles are corresponding angles where a transversal crosses parallel lines, so are congruent. That means ...

∠A = ∠B

8x -8° = 5x +25° . . . . . substitute the given expressions

3x = 33° . . . . . . . . . . . . add 8°-5x

x = 11° . . . . . . . . . . . . . . divide by 3

Then the angles are ...

8·11° -8° = 80°

6 0

2 years ago