2 years ago

−3 1/0−(−2 1/0)

Mathematics

1 answer:

lianna [129]2 years ago

4 0

I believe it's "-1" hope I helped

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Find the value of x that makes m||n

svet-max [94.6K]

7x-23+11x+23=180
4x+0=180
4x/4=180/4
X=45

Answer: x=45

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3 years ago

Two semicircles are attached to the sides of a rectangle as shown.

NeTakaya

Answer:

157 in²

Step-by-step explanation:

The area of the shape is the sum of three sections

1. Rectangle

A = 5*14 = 70 in²

2. Bigger semicircle

A = 1/2π(14/2)² = 76.96 ≈ 77 in² (rounded)

3. Smaller semicircle

A = 1/2π(5/2)² = 9.8174 ≈ 10 in² (rounded)

Total area:

70 + 77 + 10 = 157 in²

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3 years ago

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All students in Ridgewood Junior High School either get their lunch in the school cafeteria or brought it from home on Tuesday.

grigory [225]

Answer: 720

Step-by-step explanation:

5%=36

1%=36/5

1%=7.2

100%=7.2*100

100%=720 students

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3 years ago

What percent represents the part of the model shaded green. A.60% B.62.5%. C.66.7% D.75%

VLD [36.1K]

Answer:

60%

Step-by-step explanation:

3/5 = 6/10 = 60%

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2 years ago

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Suppose that the terminal side of angle alphaα lies in Quadrant I and the terminal side of angle betaβ lies in Quadrant IV. If s

melamori03 [73]

Solution :

It is given that :

$\alpha$ lies in the first quadrant.

And $\beta$ lies in the fourth quadrant.

Since, $\sin \alpha = \frac{5}{13}$ and $\cos \beta = \frac{6}{\sqrt{85}}$ (given)

$\sin \alpha = \frac{5}{13}$

$\cos \alpha = \sqrt{1-\sin^2 \alpha}$

$\cos \alpha = \frac{12}{13}$

Similarly $\cos \beta = \frac{6}{\sqrt{85}}$

$\sin \beta = \sqrt{1-\cos^2 \beta}$

$\sin \beta = \sqrt{1-\frac{36}{85}}$

$-\frac{7}{\sqrt{85}}$ (IVth quadrant)

Therefore,

$\cos (\alpha + \beta) = \cos \alpha \cos \beta - \sin \alpha \sin \beta$

$=\frac{12}{13}\times \frac{6}{\sqrt{85}}-\frac{5}{13}\times \frac{-7}{\sqrt{85}}$

$= \frac{107}{13 \sqrt{85}}$

6 0

2 years ago