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Sloan [31]
2 years ago
10

If PQRS is a rhombus,which statement must be true? Check all that apply.

Mathematics
1 answer:
Oxana [17]2 years ago
8 0

Answer: A, B, C, D

Step-by-step explanation:

A. This is true because all rhombi are parallelograms, and diagonals of a parallelogram bisect each other.

B. This is true because the diagonals of a rhombus are perpendicular.

C. This is true because diagonals of a rhombus bisect the angles from which they are drawn,

D. This is true because all sides of a rhombus are congfruent.

E. This is not always true - all rhombi are parallelograms, and adjacent angles of a parallelogram are supplementary, but not always congruent.

F. This is not always true - diagonals of a rhombus are not always congruent.

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Can I get some help? <br><br>I don't get this question in Geometry....Any help?
Dmitrij [34]
Scale Factor = 3/2
6 X (3/2) = x-2
9 = x-2
x = 11

8 X (3/2) = y + 3
12 = y+3
y = 9

6 0
4 years ago
RATS is a parallelogram. M
AysviL [449]

Answer:

Answer:

x = 12

Step-by-step explanation:

m∠S + m∠T = 180

8x + 7x = 180

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does this help?

7 0
3 years ago
Francesca is growing 1/2 inch each year. How many years will it take her to grow 6 inches?
tresset_1 [31]

Answer:

3 years.

6÷ .5= 3

edit:

brain fart! sorry!

12 years.

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4 0
3 years ago
Read 2 more answers
the amount of photos you take while on vacation varies directly with the amount of days you spend relaxing after 3 days of vacat
Sindrei [870]
I am not sure what kind of equation you want, but you could possibly use something like _ x 3 = 15
6 0
4 years ago
(X^2+x-6)(2x^2+4x) factor completely
MissTica

Answer:

2x\left(x-2\right)\left(x+3\right)\left(x+2\right)

Step-by-step explanation:

Lets go ahead and take a step by step approach to solving your problem.

First we must start by factoring x^2+x-6.

To do this, we must break the expression into groups! Although I will also provide a definition to help you understand! For ax^2+bx+c we need to find u, v > u\cdot \:v=a\cdot \:c and u+v=b which we will group into \left(ax^2+ux\right)+\left(vx+c\right).

The values we have are...a=1,\:b=1,\:c=-6 and u\cdot v=-6,\:u+v=1.

Now we need the factors of 6 which are 1, 2, 3 and 6. We also need the negative factors which you get simply by multiplying the positives by -1 or just reversing them.

Now we need to check for every two factors if u + v = 1.

\mathrm{Check}\:u=1,\:v=-6: \:u\cdot v=-6,\:u+v=-5 \Rightarrow  \mathrm{False}

\mathrm{Check}\:u=2,\:v=-3: \:u\cdot v=-6,\:u+v=-1 \Rightarrow  \mathrm{False}

\mathrm{Check}\:u=3,\:v=-2: \:u\cdot v=-6,\:u+v=1 \Rightarrow  \mathrm{True}

\mathrm{Check}\:u=6,\:v=-1: \:u\cdot v=-6,\:u+v=5 \Rightarrow  \mathrm{False}

Therefore u=3,\:v=-2.

Now we want to \mathrm{Group\:into\:}\left(ax^2+ux\right)+\left(vx+c\right). Which is \left(x^2-2x\right)+\left(3x-6\right).

Now we must factor x from x^2-2x.

Lets apply the exponent rule of a^{b+c}=a^ba^c which means x^2=xx.

Therefore we now have xx - 2x. Lets factor out the common term of x to get x(x - 2).

Now factor 3 out of 3x-6.

Rewrite 6 as 3 * 2. 3x-3\cdot \:2.

Now factor out the common term of 3 > 3\left(x-2\right).

x\left(x-2\right)+3\left(x-2\right) > Factor out the common term of x - 2 > \left(x-2\right)\left(x+3\right).

Now lets factor 2x^2+4x.

Apply the previous exponent rule of a^{b+c}=a^ba^c > x^2=xx > 2xx+4x.

Now rewrite 4 as 2 * 2 > 2xx+2\cdot \:2x.

Now factor out the common term of 2x to get 2x(x + 2).

Combine it all and we get 2x\left(x-2\right)\left(x+3\right)\left(x+2\right).

Hope this helps!

7 0
3 years ago
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