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almond37 [142]
3 years ago
9

Find x in the given figure.

Mathematics
1 answer:
professor190 [17]3 years ago
7 0

Answer:

you... didn't put anything....

Step-by-step explanation:

try asking it again, but put an image

You might be interested in
Six more than 1/5 a number is 12. What is the number
suter [353]

Answer:

Step-by-step explanation:

let x be the number

6+1/5x=12

1/5x=12-6

1/5x=6

x=5*6

x=30

check: 6+(1/5*30)=6+6=12

8 0
3 years ago
Simplify (6x2 + 11x − 3) + (2x2 − 17x − 4).
d1i1m1o1n [39]

Answer:

B. 8x^2-6x-7

Step-by-step explanation:

All you have to do is combine the like terms.

Like terms are the terms that have the same variable and same exponent.

The like terms in this equation are 6x^2 and 2x^2, 11x and -17x, and -3 and -4.

When you add 6x^2 and 2x^2, you get 8x^2

When you add 11x and -17x, you get -6x

When you add -3 and -4, you get -7.

Putting these all in order, your answer is

8x^2 - 6x - 7

8 0
3 years ago
Calculate ZQRP<br> 12 cm<br> Р<br> R<br> 18 cm
gladu [14]

Answer:

<QRP = 48.2°

Step-by-step explanation:

Given:

PR = 18 cm

QR = 12 cm

Required:

<QRP = angle R

Solution:

Reference angle = <R (<QRP)

Adjacent side length = 12 cm

Hypotenuse = 18 cm

Thus:

Cos (R) = 12/18

cos (R) = 0.6667

R = Cos^{-1}(0.6667)

R = 48.1871227

<QRP = 48.2° (nearest tenth)

8 0
3 years ago
A substance with a half life is decaying exponentially. If there are initially 12 grams of the substance and after 70 minutes th
77julia77 [94]

Answer: 233 min

Step-by-step explanation:

This problem can be solved by the following equation:

A=A_{o} e^{-kt}  (1)

Where:

A=7 g is the quantity left after time t

A_{o}=12 g is the initial quantity

t=70 min is the time elapsed

k is the constant of decay for the material

So, firstly we need to find the value of k from (1) in order to move to the next part of the problem:

\frac{A}{A_{o}}=e^{-kt}  (2)

Applying natural logarithm on both sides of the equation:

ln(\frac{A}{A_{o}})=ln(e^{-kt})  (3)

ln(\frac{A}{A_{o}})=-kt  (4)

k=-\frac{ln(\frac{A}{A_{o}})}{t}  (5)

k=-\frac{ln(\frac{7 g}{12 g})}{70 min}  (6)

k=0.00769995 min^{-1}  (7)  Now that we have the value of k we can solve the other part of this problem: Find the time t for A=2 g.

In this case we need to isolate t from (1):

t=-\frac{ln(\frac{A}{A_{o}})}{k}  (8)

t=-\frac{ln(\frac{2 g}{12 g})}{0.00769995 min^{-1}}  (9)

Finally:

t=232.697 min \approx 233 min

5 0
3 years ago
The formula for the lateral surface area of a cylinder is S=2πrhS=2πrh , where r is the radius of the bases and h is the height.
rewona [7]

Answer:

S / (2 * pi * h) =  r

Step-by-step explanation:

S = 2*pi * r * h

solve for r

Divide each side by 2* pi * h to isolate r

S / (2 * pi * h) = 2*pi * r * h/( 2 * pi * h)

S / (2 * pi * h) =  r

6 0
3 years ago
Read 2 more answers
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