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Bad White [126]
3 years ago
7

Find the area. Round to nearest tenth.

Mathematics
1 answer:
Aleksandr-060686 [28]3 years ago
8 0
Area of Triangle: 120

Area of Circle 1: <span>132.73

Area of Half-Circle 2: </span><span>39.27

120 + 132.73 + 39.27 = 292 m²</span>
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What is the area of the shape shown?
Gnesinka [82]

Answer:

B: 378 cm^2

Step-by-step explanation:

22 * 24 = 528

528 - (10 * 15) = 378

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3 years ago
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Can yall help me with this<br> WILL GIVE BRAINLIEST
dexar [7]
19 guests because each guest is $2 each according to the question.
6 0
2 years ago
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Please answer! I crossed out the ones you don’t have to complete.
Nina [5.8K]

Answer:

1. Rewriting the expression 5.a.b.b.5.c.a.b.5.b using exponents we get: \mathbf{5^3a^2b^4c}

5.  x^-6 = \frac{1}{x^6}

6. 5^{-3}.3^{-1}=\frac{1}{5^3.3^1}

7. a^{-3}b^0c^4=\frac{c^4}{a^3}

Step-by-step explanation:

Question 1:

We need to rewrite the expression using exponents

5.a.b.b.5.c.a.b.5.b

We will first combine the like terms

5.5.5.a.a.b.b.b.b.c

Now, if we have 5.5.5 we can write it in exponent as: =5^{1+1+1}=5^3

a.a as a^{1+1}=a^2

b.b.b.b as: b^{1+1+1+1}=b^4

So, our result will be:

5^3a^2b^4c

Rewriting the expression 5.a.b.b.5.c.a.b.5.b using exponents we get: \mathbf{5^3a^2b^4c}

Question:

Rewrite using positive exponent:

The rule used here will be: a^{-1}=\frac{1}{a^1} which states that if we need to make exponent positive, we will take it to the denominator.

Applying thee above rule for getting the answers:

5) x^{-6} = \frac{1}{x^6}

6) 5^{-3}.3^{-1}=\frac{1}{5^3.3^1}

7) a^{-3}b^0c^4=\frac{b^0c^4}{a^3}

We know that b^0=1 so, we get

a^{-3}b^0c^4=\frac{b^0c^4}{a^3}=\frac{c^4}{a^3}

4 0
3 years ago
Elijah opens a savings account with $60. Each week after, he deposits $25. In how many weeks will he have saved $600?
iren2701 [21]

Answer:

25w + 60= 600

25w= 540

w=21.6

Step-by-step explanation:

First make the equation

1. Subrtract 60 from each side

2. Now divide 25 on both sides

3. now you have the answer

I GOT A DECIMAL SO MAYBE ROUND OR PUT THAT EXACT ANSWER.

3 0
3 years ago
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A petrol kiosk p is 12 km due north of another petrol kiosk q. The bearing of a police station r from p is 135 degree and that f
tiny-mole [99]

Answer:

Distance between P and R is 40.15 km.

Step-by-step explanation:

From the picture attached,

Petrol kiosk P is 12 km due North of another petrol kiosk Q.

Bearing of a police station R is 135° from P and 120° from Q.

m∠QPR = 180° - 135° = 45°

m∠PQR = 120°

m∠PRQ = 180° - (m∠QPR +m∠PQR)

             = 180° - (45° + 120°)

             = 180° - 165°

             = 15°

Now we apply sine rule in ΔPQR to measure the distance between P and R.

\frac{\text{sin}(\angle QPR)}{\text{QR}}= \frac{\text{sin}(\angle PQR)}{\text{PR}}=\frac{\text{sin}\angle PRQ}{\text{PQ}}

\frac{\text{sin}(45)}{\text{QR}}= \frac{\text{sin}(120)}{\text{PR}}=\frac{\text{sin}(15)}{\text{12}}

\frac{\text{sin}(120)}{\text{PR}}=\frac{\text{sin}(15)}{\text{12}}

PR = \frac{12\text{sin}(120)}{\text{sin}(15)}

PR = 40.15 km

Therefore, distance between P and R is 40.15 km.

8 0
3 years ago
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