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2 years ago

Does anyone know this answer?

Mathematics

2 answers:

Olegator [25]2 years ago

6 0

Answer:

2x = 40°

3x = 60°

4x = 80°

Step-by-step explanation:

2x+3x+4x = 180°

9x = 180

x = 20

2x = 2(20) = 40°

3x = 3(20) = 60°

4x = 4(20) = 80°

Allushta [10]2 years ago

5 0

2x=40
3x=60
4x=80
By 2x+3x+4x=180
9x=180 divided both side by 9
X=20

Check answer by subtitling
2(20)=40
3(20)=60
4(20)=80

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Can someone help with this question

Kay [80]

Answer:

133pi

98% sure this is right

circumference in pi form is the radius time 2 and then put pi after it

Step-by-step explanation:

4 0

3 years ago

What is the distance between -8/5 and -3/5<br> A.-1<br> B.1<br> C.11/5<br> D. -11/5

Maurinko [17]

Answer:

Step-by-step explanation:

Subtract the two numbers and take the positive value ( since distance is positive)

-8/5 - (-3/5)

-8/5 + 3/5

-5/5

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3 years ago

Item 3 The Hampton family used 17,158 kilowatts of electricity last year. They used about the same amount of electricity each we

katen-ka-za [31]

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7 0

2 years ago

Given the speeds of each runner below, determine who runs the fastest.

uranmaximum [27]

Answer:

Liz

Step-by-step explanation:

lets make all speeds ft/1sec

Ron 8ft/1sec

Liz 104ft/7sec : 104/7=

14.8571428571 ft/1 sec

Katie

1 mile = 5280ft

5280ft/526sec

5280/526=

10.0380228137ft/1 sec

Emily,

1 min= 60 sec

835ft/60 sec

835/60=

13.9166666667 ft/sec

now, lets order from least to greatest

R:8ft/1sec, K:10.0380228137ft/1 sec,

E:13.9166666667 ft/sec, L:14.8571428571 ft/1 sec

Liz's speed is fastest

4 0

3 years ago

Tan^2 A/1+cot^2 A + cot^2 A/1+tan^2 A=sec^2 A cosec^2 A-3

omeli [17]

$\frac{tan^2x}{1+cot^2x}+\frac{cot^2x}{1+tan^2x}=sec^2x\ cosec^2x-3\\\\L=\frac{tan^2x(1+tan^2x)+cot^2x(1+cot^2x)}{(1+cot^2x)(1+tan^2x)}=\frac{tan^2x+tan^4x+cot^2x+cot^4x}{1+tan^2x+cot^2x+tanxcotx}\\\\=\frac{tan^2x+cot^2x+tan^4x+cot^4x}{1+tan^2x+cot^2x+1}=\frac{tan^2x+2+cot^2x+tan^4x-2+cot^4x}{tan^2x+cot^2x+2}$

$=\frac{(tanx+cotx)^2+(tan^2x-cot^2x)^2}{(tanx+cotx)^2}=\frac{(tanx+cotx)^2}{(tanx+cotx)^2}+\frac{(tan^2x-cot^2x)^2}{(tanx+cotx)^2}\\\\=1+\frac{(tanx-cotx)^2(tanx+cotx)^2}{(tanx+cotx)^2}=1+(tanx-cotx)^2\\\\=1+tan^2x-2tanx\ cotx+cot^2x=tan^2x+cot^2x+1-2\\\\=\left(\frac{sinx}{cosx}\right)^2+\left(\frac{cosx}{sinx}\right)^2-1=\frac{sin^2x}{cos^2x}+\frac{cos^2x}{sin^2x}-1=\frac{sin^4x+cos^4x}{sin^2x\ cos^2x}-1$

$=\frac{(sin^2x)^2+2sin^2x\ cos^2x+(cos^2x)^2-2sin^2x\ cos^2x}{sin^2x\ cos^2x}-1\\\\=\frac{(sin^2x+cos^2x)^2-2sin^2x\ cos^2x}{sin^2x\ cos^2x}-1=\frac{1^2-2sin^2x\ cos^2x}{sin^2x\ cos^2x}-1\\\\=\frac{1}{sin^2x\ cos^2x}-\frac{2sin^2x\ cos^2x}{sin^2x\ cos^2x}-1=\frac{1}{sin^2x}\cdot\frac{1}{cos^2x}-2-1\\\\=cosec^2x\cdot sec^2x-3=sec^2x\ cosec^2x-3=R$

3 0

3 years ago