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3 years ago

Can someone help me with this please

Mathematics

1 answer:

Annette [7]3 years ago

6 0

Answer:

The answer to your question is A.

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Please help me out please

natita [175]

Answer:

RC = 40

Step-by-step explanation:

Note that the circumcentre is equally distant from the triangle's 3 vertices.

That is : PC = RC = QC

Equate any pair and solve for x

Using RC = PC, then

5x - 15 = 3x + 7 ( subtract 3x from both sides )

2x - 15 = 7 ( add 15 to both sides )

2x = 22 ( divide both sides by 2 )

x = 11

Hence

RC = (5 × 11) - 15 = 55 - 15 = 40 units

5 0

3 years ago

A swim coach has at most $800 to buy equipment for the swim team. Each pull buoy costs $5.50 and each kickboard costs $10. Let x

MissTica

Answer:

5.50x+10y≤800

Step-by-step explanation:

The pull bouys are $5.50 each (5.50x)

The kickboards are $10 each (10y)

And the words "at most" mean less than or equal to

Hope this helps

(can I plz have a brainliest)

3 0

4 years ago

Fine length of BC on the following photo.

MrMuchimi

Answer:

$BC=4\sqrt{5}\ units$

Step-by-step explanation:

see the attached figure with letters to better understand the problem

step 1

In the right triangle ACD

Find the length side AC

Applying the Pythagorean Theorem

$AC^2=AD^2+DC^2$

substitute the given values

$AC^2=16^2+8^2$

$AC^2=320$

$AC=\sqrt{320}\ units$

simplify

$AC=8\sqrt{5}\ units$

step 2

In the right triangle ACD

Find the cosine of angle CAD

$cos(\angle CAD)=\frac{AD}{AC}$

substitute the given values

$cos(\angle CAD)=\frac{16}{8\sqrt{5}}$

$cos(\angle CAD)=\frac{2}{\sqrt{5}}$ ----> equation A

step 3

In the right triangle ABC

Find the cosine of angle BAC

$cos(\angle BAC)=\frac{AC}{AB}$

substitute the given values

$cos(\angle BAC)=\frac{8\sqrt{5}}{16+x}$ ----> equation B

step 4

Find the value of x

In this problem

$\angle CAD=\angle BAC$ ----> is the same angle

equate equation A and equation B

$\frac{8\sqrt{5}}{16+x}=\frac{2}{\sqrt{5}}$

solve for x

Multiply in cross

$(8\sqrt{5})(\sqrt{5})=(16+x)(2)\\\\40=32+2x\\\\2x=40-32\\\\2x=8\\\\x=4\ units$

$DB=4\ units$

step 5

Find the length of BC

In the right triangle BCD

Applying the Pythagorean Theorem

$BC^2=DC^2+DB^2$

substitute the given values

$BC^2=8^2+4^2$

$BC^2=80$

$BC=\sqrt{80}\ units$

simplify

$BC=4\sqrt{5}\ units$

7 0

3 years ago

After the translation, where is A located?

vovikov84 [41]

Answer:

(-3, 13)

Step-by-step explanation:

The transformation that moves a point 4 left and 8 up is ...

(x, y) ⇒ (x -4, y +8)

The transformation that reflects a point across the y-axis is ...

(x, y) ⇒ (-x, y)

Applied after the translation, the transformation of ∆ABC becomes ...

(x, y) ⇒ (-(x -4), y +8) = (4 -x, y +8)

Then point A gets moved to ...

A(7, 5) ⇒ A'(4 -7, 5 +8) = (-3, 13)

7 0

2 years ago

Please help with this

Karolina [17]

x = -10 that is the answer your welcome

5 0

3 years ago