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3 years ago

9 minutes left to finish this!! I need help!

Mathematics

2 answers:

lbvjy [14]3 years ago

6 0

Answer:

p = 4

j = 21

Step-by-step explanation:

Joey = j

Pat = p

j = p + 17

(j+6) = 2*(p + 6) + 7 Simplify this. Remove the brackets.

j + 6 = 2p + 12 + 7 combine like terms

j + 6 = 2p + 19 Subtract 6 from both sides

j +6-6 = 2p +19-6

j = 2p + 13

================

Equation j = 2p + 13 and j = p + 17

2p + 13 = p + 17 Subtract p from both sides

2p-p+13 =p-p + 17

p + 13 = 17 Subtract 13 from both sides

p = 17-13

p = 4

============

j = p + 17

j = 4 + 17

j = 21

Leya [2.2K]3 years ago

5 0

Answer:

Joey is 21; Pat is 4

Step-by-step explanation:

The problem statement supports two equations in Joey's age (j) and Pat's age (p):

j - p = 17

(j +6) -2(p +6) = 7

Subtracting the second equation from the first, we have ...

(j -p) -((j +6) -2(p +6)) = (17) -(7)

p +6 = 10 . . . . . simplify

p = 4 . . . . . . . . . subtract 6

J = 17 +4 = 21

Joey is 21; Pat is 4.

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-5x+10<30 plz help Hurry plz...

Phoenix [80]

Answer:

x= -4

Step-by-step explanation:

Step 1- Subtract 10 to both sides.

-5x+10< 30

-10 -10

-5x< 20

Step 2- Divide both sides by -5.

-5x< 20

-5 -5

x= -4

Check 

-5(-4)+10< 30

-20+10< 30

-10< 30

Hope this helps! :)

5 0

3 years ago

Please help me i really need it i know that it is a lot but i could really use it

k0ka [10]

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6. 25/121?

7. 1/256

8. -1/531441?

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10. 1/12t^3

11. 25/b^4

12. 7/r^7

13. 4x^5/y^3

14. 3g^3

I hope this is all right, if it's not I apologize.

7 0

3 years ago

Arrange the steps in the correct order to solve: sin 3pi/4 cos pi/12

Dmitry [639]

$\bf \textit{Product to Sum Identities}
\\\\
sin(\alpha)cos(\beta)=\cfrac{1}{2}[sin(\alpha+\beta)\quad +\quad sin(\alpha-\beta)]\\\\
-------------------------------\\\\
sin\left( \frac{3\pi }{4} \right)cos\left( \frac{\pi }{12} \right)\implies \cfrac{1}{2}\left[ sin\left( \frac{3\pi }{4}+\frac{\pi }{12} \right) ~~+~~sin\left( \frac{3\pi }{4}-\frac{\pi }{12} \right)\right]$

$\bf \cfrac{1}{2}\left[ sin\left( \frac{9\pi+\pi }{12} \right) ~~+~~sin\left( \frac{9\pi-\pi }{12} \right)\right]
\\\\\\
\cfrac{1}{2}\left[ sin\left( \frac{10\pi }{12} \right) ~~+~~sin\left( \frac{8\pi }{12} \right)\right]\implies 
\cfrac{1}{2}\left[ sin\left( \frac{5\pi }{6} \right) ~~+~~sin\left( \frac{2\pi }{3} \right)\right]
\\\\\\
\cfrac{1}{2}\left[\cfrac{1}{2}~~+~~\cfrac{\sqrt{3}}{2} \right]\implies \cfrac{1}{2}\left[ \cfrac{1+\sqrt{3}}{2} \right]\implies \cfrac{1+\sqrt{3}}{4}$

5 0

4 years ago

Please help meeeee I need correct asnwers

ycow [4]

Answer:

diameter = 12 cm

radius = 6cm

circumference = 12 $\pi$ cm ≈ 37.68 cm

area = 36 $\pi$ cm² ≈ 113.04 cm²

Step-by-step explanation:

r = radius

d = diameter

C = circumference

A = area

The diagram shows a circle with a diameter of 12 cm

Diameter

d = 12 cm

Radius

r = d/2

⇒ r = 12/2

⇒ r = 6 cm

Circumference

C = $\pi$ d

⇒ C = 12 $\pi$ cm

Area

A = $\pi$ r²

⇒ A = $\pi$ x 6²

⇒ A = 36 $\pi$ cm²

If you use $\pi$ = 3.14 then:

C = 37.68 cm

A = 113.04 cm²

4 0

2 years ago

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Serga [27]

Answer:

A, B, C, and E

Step-by-step explanation:

SSA doesn't work because this leaves room for two possible answers in some situations.

The other postulate, HL (hypotenuse-leg) works for right triangles; it's basically a short cut for SSS postulate because if the hypotenuse and leg of a right triangle are congruent to the hypotenuse and leg on another, the other side also has to be congruent.

3 0

3 years ago

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