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1 year ago

Find the area of the parallelogram.

Mathematics

2 answers:

Shalnov [3]1 year ago

8 0

Answer:

35 cm²

Step-by-step explanation:

This is because the formula of the area of a parallelogram is;

area = (base) x (height)

since the base is 5, and the height is 7;

area = (5) x (7)

area = 35

<em>Hope this helps!</em>

zubka84 [21]1 year ago

6 0

Answer:

$\boxed{\sf{A. \quad 35cm^2}}$

Step-by-step explanation:

The formula of area of the parallelogram is used to find the area of the parallelogram.

<u>Area of the parallelogram formula:</u>

$\Longrightarrow: \sf{A=B*H}$

B stands for "base".

H stands for "height".

The base is 5.
The height is 7.

<u />

<u>Multiply.</u>

<u />

<u>Solve.</u>

5*7=35

<u>Therefore, the area of the parallelogram is 35cm², which is our answer.</u>

I hope this helps! Let me know if you have any questions.

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I would like some help. THanks.

ikadub [295]

Answer:

8.5

Step-by-step explanation:

Volume= base*height

V= 2 1/8*4

V= 8.5

7 0

2 years ago

What is a correct congruence statement for the triangles shown? Enter your answer in the box. △QDJ≅△ A triangle Q D J. The b

almond37 [142]

Answer: The correct congruence statement is $\triangle QDJ\cong \triangle MCW$ .

Explanation:

It is given that A triangle Q D J. The base D J is horizontal and side Q D is vertical. Another triangle M C W is made. The base C W and side M C are neither horizontal nor vertical. Triangle M C W is to the right of triangle Q D J.

The sides Q D and M C are labeled with a single tick mark. The sides D J and C W are labeled with a double tick mark. The sides Q J and M W are labeled with a triple tick mark.

Draw two triangles according to the given information.

From the figure it is noticed that

$QD\cong MC$

$DJ\cong CW$

$QJ\cong MW$

So by SSS rule of congruence we can say that $\triangle QDJ\cong \triangle MCW$ .

4 0

2 years ago

Read 2 more answers

Prove that if a and b are positive integers whose sum is a prime p, their greatest common divisor is 1.

Tasya [4]

Proof by contadiction.

Let assume there exist such positive integers $a$ and $b$ whose sum is a prime number $p$ , that their greatest common divisor is greater than 1.

$a,b\in\mathbb{Z^+}\\d=\text{gcd}(a,b)>1\\\\a=de\\b=df\\e,f\in\mathbb{Z^+}\\\\a+b=p\\de+df=p\\d(e+f)=p\\$

Since $p$ is a prime number and $d>1$ , then $d=p \wedge e+f=1$ , but we assumed earlier that $e,f\in\mathbb{Z^+}$ , and there are no two positive integers that sum up to 1.