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2 years ago

What is the answer for 16.9+12.1=

Mathematics

1 answer:

marishachu [46]2 years ago

3 0

16.9+12.1=29.0
I hope That I helped and I got my math right! You can just use a calculator!

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If the square below has an area of 50, what is the value of x

pishuonlain [190]

Answer:

Step-by-step explanation:

A= L²

50 = L² (do opposite operation to find L)

L = 7.07

x²= L²+ L²

x²= 7.07² + 7.07²

x² = 100 ( do opposite operation to make x the subject)

x = 10

Note: opposite operation of square is square root

4 0

2 years ago

What is 3/4 of 4/5. Plz help

Dmitriy789 [7]

Answer:

3/5

Step-by-step explanation:

We want to find 3/4 of 4/5.

The keyword <em>of</em> tells us that we need to multiply. Therefore:

$\frac{3}{4}\times\frac{4}{5}$

Multiply:

$=12/20=3/5$

So, 3/4 of 4/5 is 3/5.

We can confirm this by converting them to decimals.

4/5 is 0.8.

3/4 is 0.75.

0.75 or 75% of 0.8 is indeed 3/5 or 0.6.

6 0

2 years ago

The perimeter of an isosceles triangle is 71 centimeters. The measure of one of the sides is 22 centimeters. What are all the po

denis23 [38]

An isosceles triangle has two equal side lengths and the other side length is different. Therefore, possible measures of the other two sides are either
22 cm and 27 cm, or 22 cm and 24.5 cm.

4 0

3 years ago

A car needs fixing and Abe can fix it for $70 per hour with a $130 part, but Gabe can fix it for $80 an hour with a $40 part. Ho

Shtirlitz [24]

For the same amount set up two equations set equal to each other.
$70h + 130 = 80h + $40
Subtract 80h from both sides
-10h + 130 = 40
subtract 130 from both sides
-10h = -90 divide both sides by -10
h = 9 hours

CHECK
70(9) + 130 = $760
80(9) + 40 = $760

8 0

3 years ago

Read 2 more answers

The function f(x) = StartRoot negative x EndRoot is represented by the graph. Which statement is correct?

Svetlanka [38]

Answer:

D. The range of the function is all real numbers greater than or equal to 0.

Step-by-step explanation:

The graph of the function $y=\sqrt{-x}$ is shown in attached diagram.