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2 years ago

Can you please help I am getting graded:) show work too please

Mathematics

1 answer:

Oliga [24]2 years ago

4 0

Answer: A - 33/100

Step-by-step explanation:

You have to multiply the first fraction by 10 to make 100 the least common denominator - 3/10(10) = 30/100

Add 30/100+3/100 = 33/100

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Determine the product of (46.2 × 10^-1) ⋅ (5.7 × 10^–6). Write your answer in scientific notation.

shepuryov [24]

Step-by-step explanation:

here's the answer to your question

5 0

3 years ago

Find the surface area of a square pyramid with base edges of 5 cm and a slant height of 3 cm.

ch4aika [34]

Answer:

Surface area of square pyramid = $55cm^{2}$

Step-by-step explanation:

Let, b is the base edges and s is the slant height

According to given condition:

b = 5 cm

s = 3 cm

Find out:

Surface area of square pyramid = ?

Formula:

Surface area of square pyramid = $b^{2} + 2bs$

Solution:

Surface area of square pyramid = $b^{2} + 2bs$

Surface area of square pyramid = $5^{2} + 2(5)(3)$

Surface area of square pyramid = $(25+ 30 ) cm^{2}$

Surface area of square pyramid = $55cm^{2}$

3 0

3 years ago

At 2 PM, a thermometer reading 80°F is taken outside where the air temperature is 20°F. At 2:03 PM, the temperature reading yiel

Alexeev081 [22]

Use Law of Cooling:
$T(t) = (T_0 - T_A)e^{-kt} +T_A$
T0 = initial temperature, TA = ambient or final temperature

First solve for k using given info, T(3) = 42
$42 = (80-20)e^{-3k} +20 \\ \\ e^{-3k} = \frac{22}{60}=\frac{11}{30} \\ \\ k = -\frac{1}{3} ln (\frac{11}{30})$

Substituting k back into cooling equation gives:
$T(t) = 60(\frac{11}{30})^{t/3} + 20$

At some time "t", it is brought back inside at temperature "x".
We know that temperature goes back up to 71 at 2:10 so the time it is inside is 10-t, where t is time that it had been outside.
The new cooling equation for when its back inside is:
$T(t) = (x-80)(\frac{11}{30})^{t/3} + 80 \\ \\ 71 = (x-80)(\frac{11}{30})^{\frac{10-t}{3}} + 80$
Solve for x:
$x = -9(\frac{11}{30})^{\frac{t-10}{3}} + 80$
Sub back into original cooling equation, x = T(t)
$-9(\frac{11}{30})^{\frac{t-10}{3}} + 80 = 60 (\frac{11}{30})^{t/3} +20$
Solve for t:
$60 (\frac{11}{30})^{t/3} +9(\frac{11}{30})^{\frac{-10}{3}}(\frac{11}{30})^{t/3} = 60 \\ \\ (\frac{11}{30})^{t/3} = \frac{60}{60+9(\frac{11}{30})^{\frac{-10}{3}}} \\ \\ \\ t = 3(\frac{ln(\frac{60}{60+9(\frac{11}{30})^{\frac{-10}{3}}})}{ln (\frac{11}{30})}) \\ \\ \\ t = 4.959$

This means the exact time it was brought indoors was about 2.5 seconds before 2:05 PM

8 0

3 years ago

Find the missing length of the triangle. 10 in. 16 26 in.

Yanka [14]

Answer:

Step-by-step explanation:

26x26=676

10x10=100

676/100=576

The square root of 576 is 24.

8 0

2 years ago

What is the slope of the line? A. 3 B. -3 C. 1/3 D. -1/3

Inessa05 [86]

Answer:

B. -3

Step-by-step explanation:

Starting from the y-intercept of [0, -2], you do rise\run by either moving three blocks north over one block west, or three blocks south over one block east [west and south are negatives].

I am joyous to assist you anytime.

3 0

3 years ago