All categories

3 years ago

Two triangles with the same corresponding side lengths will be congruent.

Mathematics

1 answer:

OleMash [197]3 years ago

4 0

If Its A True or False Question, It'd Be "True"

You might be interested in

Sheila eats 3/4 of a bag of carrots each week, how many bags of baby carrots does she eat in 6 weeks? Write in simplest form

Alla [95]

3/4 x 6 = 18/4 = 4 1/2

she eats 4 1/2 bags of carrots in 6 weeks

8 0

2 years ago

Read 2 more answers

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

2 years ago

What is the difference between 100 and 33?

Romashka-Z-Leto [24]

Answer:

Step-by-step explanation:

100 - 33 = 67

The difference between 100 and 33 is 67.

8 0

1 year ago

Read 2 more answers

Evaluate:<br> -9 + 25 - (-10) + (-6)

Minchanka [31]

-9 is a negative number, so plus 25, it would be 16. The minus sign and the negative sign of -10 cancel out, so it is plus 10, ending up with 26. Finally, the + and - becomes negative (positive times negative equals negative), so it is -6 and you’ll end up with a final answer of 20.

4 0

2 years ago

Read 2 more answers

I’m confused on where to shade these equations? I can do the first 2 but then the last 2 throw me off... I have to shade the are

tigry1 [53]

I’m not sure the answer but go into a webpage called Desmos and it will show you

4 0

2 years ago