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Ganezh [65]
3 years ago
12

If a base of a rectangle is 36cm and the area is 845cm2 what is the height of the rectangle

Mathematics
1 answer:
sveta [45]3 years ago
8 0
\sf~A=bh

Plug in what we know:

\sf845=36h

Divide 36 to both sides:

\sf~h\approx23.47
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May I please receive help
mafiozo [28]

Hello mate ☺️,

We know that m < 2 = 96°, as,

m < 5 = 96°(Alternate angles are equal)

Now we know that m < 5 = 96°, so,

→ m < 5 + m < 8 = 180°(Linear Pair angles are supplementary)

→ 96° + m < 8 = 180°

→ m < 8 = (180 - 96)°

→ m < 8 = 84°

Therefore, <u>m<5 = 96°</u> & <u>m<8=84°</u>

✍️ <em>By </em><em>Benjemin</em> ☺️

3 0
2 years ago
Trapezoid ABCD has an area of 525 cm squared. If height AB = 21 cm and BC = 37 cm, what is the measure of AD?​
Annette [7]

Answer:

AD = 13

Step-by-step explanation:

Area of trapezoid ABCD

=  \frac{1}{2} (AB + AD) \times AB \\  \\  525=  \frac{1}{2} (37+ AD) \times21\\  \\  525 \times 2=  (37+ AD) \times21 \\  \\ 1050 = 21 \times 37 + 21AD \\  \\ 1050  = 777+ 21AD \\  \\  21AD  = 1050 - 777 \\  \\ 21AD  = 273 \\  \\ AD  =  \frac{273}{21}  \\  \\ AD  = 13

8 0
3 years ago
How do you do this? Please help
PtichkaEL [24]

Answer:

PO=86.2\\ OM=23.8\\

Step-by-step explanation:

To preface, your figure is going to be a line segment, with O as your midpoint, in between points P & M.

With that being said:

PO+OM=PM

Identify your values:

PO=7y+12\\OM=3y-8\\PM=110

Substitute the values into the first equation:

7y+12+3y-8=110

Combine like terms:

10y+4=110

Subtract 4 from both sides of the equation:

10y=106

Divide by the coefficient of y, which is 10:

y=10.6

Substitute 10.6 for y in segments PO & OM:

PO=7(10.6)+12

OM=3(10.6)-8

PO=74.2+12

OM=31.8-8

Solve:

PO=86.2

OM=23.8

Check your answers by substituting:

PO+OM=PM

86.2+23.8=110

8 0
4 years ago
60 = −2(−10 − 4x)<br> what is the answer show step by step
Papessa [141]
That’s the answer with steps

8 0
3 years ago
Read 2 more answers
(1) Factor the GCF out of the trinomial on the left side of the equation. (2 points: 1 for the GCF, 1 for the trinomial) 2x²+6x-
Tanya [424]

Step-by-step explanation:

(1) Factor the GCF out of the trinomial on the left side of the equation. (2 points: 1 for the GCF, 1 for the trinomial) 2x²+6x-20=0

2x²+6x-20

2(x²+3x-10)

the factors are 2 and (x²+3x-10)

(2) Factor the polynomial completely. (4 points: 2 point for each factor)

2(x²+3x-10)

2(x²-2x+5x-10)

2(x(x-2) + 5(x-2))   group like terms

2(x+5)(x-2)

(3) If a product is equal to zero, we know at least one of the factors must be zero. And the constant factor cannot be zero. So set each binomial factor equal to 0 and solve for x, the width of your project. (2 points: 1 point for each factor)

constant = 2  cannot be zero

the other factors are (x+5) and (x-2)

(x+5)=0 => x= -5

or

(x-2)=0 => x=2

(4) What are the dimensions of your project? Remember that the width of your project is represented by x. (2 points: 1 point for each dimension)

thank you so much, sorry if it's a little confusing!!  

(it is indeed confusing, because physical dimensions cannot be negative)

The dimensions of the project (assumed a rectangle) are +2 and -5

3 0
4 years ago
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