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3 years ago

If 3t+2=7, then t=? Answer choices: A.-9 B.3/5 C.5/3 D.9

Mathematics

2 answers:

lions [1.4K]3 years ago

5 0

3t+2=7, 3t=5, t= 5/3, so the answer is C. 5/3

Butoxors [25]3 years ago

4 0

The answer to you question would be c. 5/3

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Find the missing values in each row. Use the percent equation. A: Principal: $100 Interest rate: 5% Time in years: 3 Interest ea

ANEK [815]

Answer:

A.) 15

B.) 1 year

C.) $50

D.) 1.5%

Step-by-step explanation:

The percentage equation goes thus:

Interest earned = principal * rate * time

Principal: $100 Interest rate: 5% Time in years: 3 Interest earned

Interest earned = $100 * 0.05 * 3

Interest earned = $15

B.)

Principa;" $500 Interest rate: 4% Time in years: interest earned: $20

Interest earned = principal * rate * time

20 = 500 * 0.04 * time

20 = 20 * time

Time = 20 / 20 = 1

Time = 1 year

C.)

Principal: Interest rate:10% Time in years: 7 Interest earned:$35

Interest earned = principal * rate * time

$35 = principal * 0.1 * 7

$35 = principal * 0.7

Principal = $35 / 0.7

Principal = $50

D.)

Principal: $200 Interest rate: Time in years:2 Interest earned: $6

Interest earned = principal * rate * time

$6 = $200 * rate * 2

$6 = $400 * rate

Rate = $6/$400

Rate = 0.015 = 1.5%

7 0

3 years ago

Evaluate f(x - 2) when f(x) = 4x² - 4x – 2

ivann1987 [24]

Answer:

Step-by-step explanation:

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3 years ago

Find the number b such that the line y = b divides the region bounded by the curves y = 36x2 and y = 25 into two regions with eq

Gemiola [76]

Answer:

b = 15.75

Step-by-step explanation:

Lets find the interception points of the curves

36 x² = 25

x² = 25/36 = 0.69444

|x| = √(25/36) = 5/6

thus the interception points are 5/6 and -5/6. By evaluating in 0, we can conclude that the curve y=25 is above the other curve and b should be between 0 and 25 (note that 0 is the smallest value of 36 x²).

The area of the bounded region is given by the integral

$\int\limits^{5/6}_{-5/6} {(25-36 \, x^2)} \, dx = (25x - 12 \, x^3)\, |_{x=-5/6}^{x=5/6} = 25*5/6 - 12*(5/6)^3 - (25*(-5/6) - 12*(-5/6)^3) = 250/9$

The whole region has an area of 250/9. We need b such as the area of the region below the curve y =b and above y=36x^2 is 125/9. The region would be bounded by the points z and -z, for certain z (this is for the symmetry). Also for the symmetry, this region can be splitted into 2 regions with equal area: between -z and 0, and between 0 and z. The area between 0 and z should be 125/18. Note that 36 z² = b, then z = √b/6.

$125/18 = \int\limits^{\sqrt{b}/6}_0 {(b - 36 \, x^2)} \, dx = (bx - 12 \, x^3)\, |_{x = 0}^{x=\sqrt{b}/6} = b^{1.5}/6 - b^{1.5}/18 = b^{1.5}/9$