All categories

4 years ago

What is the sum of -7 and its additive inverse?

Mathematics

1 answer:

Virty [35]4 years ago

3 0

Answer:

The sum of a number and its additive inverse is equal to 0.

Step-by-step explanation:

You might be interested in

The prime factors of a composite number must be less than a square root of the composite number. True or False?

Oksi-84 [34.3K]

Answer: It depends on what you're asking. If it can be equal, then yes. If it must be less, then no.

Step-by-step explanation:

Every composite number has a square root equal or less than its square root.

7 0

3 years ago

I need help ASAP could someone help with this?

Varvara68 [4.7K]

=) sorry. but i do not know this. i hope somebody helps you out> =)

4 0

3 years ago

Help!!!Can someone help me with my math ttas fast!!

MAVERICK [17]

The answer to the question is boneless

4 0

4 years ago

Clayton is shopping for a new shower. He is comparing the water usage of cach as he is shopping .

Jlenok [28]

What else ? Is that it ?

7 0

3 years ago

Sinx = 1/2, cosy = sqrt2/2, and angle x and angle y are both in the first quadrant.

Leviafan [203]

Answer:

Option D. 3.73

Step-by-step explanation:

we know that

$tan(x+y)=\frac{tan(x)+tan(y)}{1-tan(x)tan(y)}$

and

$sin^{2}(\alpha)+cos^{2}(\alpha)=1$

step 1

Find cos(X)

we have

$sin(x)=\frac{1}{2}$

we know that

$sin^{2}(x)+cos^{2}(x)=1$

substitute

$(\frac{1}{2})^{2}+cos^{2}(x)=1$

$cos^{2}(x)=1-\frac{1}{4}$

$cos^{2}(x)=\frac{3}{4}$

$cos(x)=\frac{\sqrt{3}}{2}$

step 2

Find tan(x)

$tan(x)=sin(x)/cos(x)$

substitute

$tan(x)=1/\sqrt{3}$

step 3

Find sin(y)

we have

$cos(y)=\frac{\sqrt{2}}{2}$

we know that

$sin^{2}(y)+cos^{2}(y)=1$

substitute

$sin^{2}(y)+(\frac{\sqrt{2}}{2})^{2}=1$

$sin^{2}(y)=1-\frac{2}{4}$

$sin^{2}(y)=\frac{2}{4}$

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

step 4

Find tan(y)

$tan(y)=sin(y)/cos(y)$

substitute

$tan(y)=1$

step 5

Find tan(x+y)

$tan(x+y)=\frac{tan(x)+tan(y)}{1-tan(x)tan(y)}$

substitute

$tan(x+y)=[1/\sqrt{3}+1}]/[{1-1/\sqrt{3}}]=3.73$

7 0

3 years ago