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2 years ago

Find the area of the parallelogram

Mathematics

2 answers:

inn [45]2 years ago

8 0

Hello from MrBillDoesMath!

Answer:

32 ft^2

Discussion:

You solved it yourself. The area of a parallelogram of base "b" and height (altitude) "h" is b*h. In this case

Area = 8 * 4 = 32 ft^2

Thank you,

MrB

ahrayia [7]2 years ago

5 0

Answer:

A = 32

Step-by-step explanation:

You already wrote how to do it on your paper.

Since A = b x h, A = 8 x 4.

So, we multiply 8 x 4.

A = 32

Hope this helps! I need brainliest to level up!

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What is (8 x 10) + (6x1) + (9 x 1/10) + (5x1/1000) in standard form?

hjlf

I believe if you simply everything, you will get 17381 over 20 or 17381/20

4 0

2 years ago

Please answer my question bcz I can’t understand

RUDIKE [14]

Please send more context as of the "green numbers".

For the multiplication,

i 49 x 10 = 490
490 ÷ 10 = 49

ii 2.3 ÷ 10 = 0.23
0.23 x 10 = 2.3

iii 0.034 x 1000 = 34
34 ÷ 1000 = 0.034

iv 876 ÷ 100 = 8.76
8.76 x 100 = 876

Hope this helps :)

3 0

2 years ago

Let log base aU=X and log base aV=Y, then a to the x power =? and a to y power =?

Andrews [41]

Answer:

${ \bf{ log_{a}(U) = x}} \\ { \boxed{ \tt{ {a}^{x} = U}}} \\ \\ { \bf{ log_{a}(V) = y}} \\ { \boxed{ \tt{ {a}^{y} = V }}}$

3 0

3 years ago

Please help me

morpeh [17]

Answer:

No, because the cone has a smaller volume than the ice cream

Step-by-step explanation:

V=4

3πr3=4

3·π·33≈113.09734

Volume of scoop: 113.1 cm^2

V=πr2h

3=π·2.52·10

3≈65.44985

Volume of cone: 65.45 cm^2

65.45 < 113.1

6 0

2 years ago

Trapezoid QRST has two right angles. A 5-in.altitude can be drawn dividing QRDT into a rectangle and an isosceles right triangle

Vika [28.1K]

Answer:

The area of the trapezoid is 57.5 square inches

Step-by-step explanation:

we know that

The trapezoid QRST can be divided into a rectangle QRDT and an isosceles right triangle RSD

see the attached figure to better understand the problem

step 1

The area of rectangle is given by the formula

$A=((QR)(RD)$

we have

$RD=5\ in$ ----> altitude

$QR=9\ in$

substitute

$A=((9)(5)=45\ in^2$

step 2

Find the area of the isosceles right triangle

The area of triangle is given by the formula

$A=\frac{1}{2}(RD)(DS)$

we have

$RD=DS=5\ in$ ---> because is an isosceles triangle

substitute

$A=\frac{1}{2}(5)(5)=12.5\ in^2$

step 3

Adds the areas

$A=45+12.5=57.5\ in^2$

3 0

3 years ago

Read 2 more answers

Find the area of the parallelogram​

Find the area of the parallelogram