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antiseptic1488 [7]

3 years ago

8

Evaluate. 5 • [(14 + 3 • 30) + 120] + 103 – 10

2 answers:

xz_007 [3.2K]3 years ago

5 0

We will use the PEMDA method.
<span>5 • [(14 + 3 • 30) + 120] + 103 – 10
</span> first, add what's in the parenthesis
5 • [(17 <span>• 30) + 120] + 103 - 10
</span> now let's multiply what's in the parenthesis
5 <span>• [(510) + 120] + 103 - 10
</span> continue to simplify what is in the parenthesis
5 <span>• [630] + 103 - 10
</span>
now let's multiply
3150 + 103 - 10

no division? let's subtract!

3150 + 93

last step (:
3243 seems to be the answer

natali 33 [55]3 years ago

4 0

Simplify the expression.<span>1213</span>

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Solve.........................

neonofarm [45]

Answer:

Step-by-step explanation:

C = 2 pi * r

r = d/2

r = 6/2 = 3

C = 2 * 3.14 * 3

C = 18.85 km

6 0

4 years ago

Someone please help, having trouble with this

Dennis_Churaev [7]

Answer:

C

Step-by-step explanation:

it describes how much the shown graph extends left and right.

Note that 4 is included, but -2 itself not. that's why it's empty and A isn't correct.

B and D are completely inapplicable

pls brainliest

8 0

3 years ago

X^2-6=2-18x solve by completing the square

AnnyKZ [126]

X^1=-9[89],x^2=-9+[89]

6 0

3 years ago

Average Price A stock analyst plots the price per share of a certain common stock as a function of time and finds that it can be

mixer [17]

Answer:

42.49 unit price

Step-by-step explanation:

The average price function, S(ave), will be an integral of the price function over the period of time (6 years), that is

S(ave) = $1/6*\int\limits^6_0 {S(t)} \, dt$

= $1/6*\int\limits^6_0 {37+6e^-0.03t} \, dt$

Solving the integral, we have

S(ave) = 1/6*(37t - (6e^-0.03t)/0.03) with the limits being from t = 0 to t = 6

Hence, we have

S(ave) = $37 - \frac{200}{6}*(e^(-0.03*6) - 1)$

This resolves to 42.49 unit price

8 0

3 years ago

The height of a parallelogram is 6.5 cm.

o-na [289]

Area of parallelogram = base × height

Area = 6.5 × 13 = 84.5 cm²

8 0

4 years ago

Read 2 more answers