All categories

2 years ago

Please help geometry question

Mathematics

2 answers:

Anuta_ua [19.1K]2 years ago

8 0

Answer:

line PQ≅line SU

Step-by-step explanation:

∆PQR ∆UST

P corresponds to U

Q corresponds to S

R corresponds to T

line PQ to line US

line QR to line ST

line PR to line UT

Stella [2.4K]2 years ago

6 0

Answer:

D) QR≅ST

The answer follows from basic AAS congruences

You might be interested in

There were 324 adults surveyed. Among the participants, the mean number of hours of sleep each night was 7.5 and the standard de

kirill [66]

0.17 sense you have to round it to the nearest hundreth

5 0

2 years ago

Read 2 more answers

Q: Marci wants to prove that the diagonals of

Trava [24]

Answer: Option B

She should use the fact that the

opposite sides of a parallelogram are

congruent and then use the

Pythagorean theorem.

We cannot use the diagonals of square property because this is a rectangle, opposite angles will also not work, and we cannot use the diagonal property because thats what we have to prove.

Must click thanks and mark brainliest

7 0

2 years ago

Compare the decimals <, >, or = 0.73 ___ 0.7300

GalinKa [24]

The decimals are equal.

8 0

2 years ago

Read 2 more answers

Which of the following functions is graphed below?

Pie

Answer:

Step-by-step explanation:

C is the solution

7 0

2 years ago

Read 2 more answers

PLEASE HELP ASAP

maria [59]

Answer:

First option: $3x^2-5x=-8$

Second option: $2x^2=6x-5$

Fourth option: $-x^2-10x=34$

Step-by-step explanation:

Rewrite each equation in the form $ax^2+bx+c=0$ and then use the Discriminant formula for each equation. This is:

$D=b^2-4ac$

1) For $3x^2-5x=-8$ :

$3x^2-5x+8=0$

Then:

$D=(-5)^2-4(3)(8)=-71$

Since $D$ this equation has no real solutions, but has two complex solutions.

2) For $2x^2=6x-5$ :

$2x^2-6x+5=0$

Then:

$D=(-6)^2-4(2)(5)=-4$

Since $D$ this equation has no real solutions, but has two complex solutions.

3) For $12x=9x^2+4$ :

$9x^2-12x+4=0$

Then:

$D=(-12)^2-4(9)(4)=0$

Since $D=0$ this equation has one real solution.

4) For $-x^2-10x=34$ :

$-x^2-10x-34=0$

Then:

$D=(-10)^2-4(-1)(-34)=-36$

Since $D$ , this equation has no real solutions, but has two complex solutions.

3 0

3 years ago