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Vedmedyk [2.9K]
3 years ago
15

I need help with my home work please

Mathematics
1 answer:
shepuryov [24]3 years ago
8 0

Answer:

I WISH I COULD HELP BUT THAT IS SOME HARD WORK PLUS

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a parallelogram has an area of 288 ft squared. If the base of the parallelogram is 24 ft, what is the height of the parallelogra
labwork [276]

Answer:H=12Ft

Base:24

Area:288

A=bh

h=A/b=288/24=12ft

7 0
3 years ago
Consider the expression below.
qwelly [4]

Answer:

x = - 9, x = - 4

Step-by-step explanation:

Set (x +)(x + 9) equal to zero, that is

(x + 4)(x + 9) = 0

Equate each factor to zero and solve for x

x + 4 = 0 ⇒ x = - 4

x + 9 = 0 ⇒ x = - 9

Thus

x = - 9, x = - 4

5 0
3 years ago
A regular hexagon has a side length of 6 units.
valkas [14]

Answer:

Step-by-step explanation:

7 0
3 years ago
Read 2 more answers
What is the product of (5 m Superscript negative 2 Baseline) (2 m Superscript negative 3 Baseline)?
uysha [10]

Answer:

C ) 10/m⁵

Step-by-step explanation:

In this question, we have to find the product of two terms

First term = 5 m⁻²

Second term = 2 m⁻³

Product of both terms = (5 m⁻²) · (2 m⁻³)

The simple way of calculating product of these two term which involves constants as well as variable is that we multiply the constants together:

5·2 = 10

And we multiply the variables. According to the rules of multiplication, when 2 same constants or variables are multiplied, we can just add their powers to get the result.

(m⁻²) · (m⁻³) = m⁻²⁻³ = m⁻⁵

The product of both terms can be written as:

(product of constants)(product of variables) = (10)(m⁻⁵) = 10/m⁵

So the correct option is C

7 0
4 years ago
Read 2 more answers
Two vertical posts stand side by side. One post is 8 feet tall and the other is 17 feet tall. If a 24 foot wire is stretched bet
lianna [129]

Answer:

The distance between the posts is 3\sqrt{55} feet.

Step-by-step explanation:

In the attached figure, let DB and AC be the posts and let AD be the wire attached to the top of the posts.

BC is the distance between the posts and we need to find BC.

Note that BC = DE

Also, EC = DB = 8m and AE = AC - EC = 17 - 8 = 9 m.

Now, in the right triangle ADE,

AD^{2} =AE^{2} +DE^{2}

24^{2} =9^{2} +DE^{2}

DE^{2} =24^{2} -9^{2}

= (24+9) (24-9)

= 33(15)

= (3 × 11) × (3 × 5)

DE=3\sqrt{55}

Since DE = BC,

BC=3\sqrt{55}

Hence, the distance between the posts is 3\sqrt{55} feet.



7 0
3 years ago
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