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3 years ago

The legs of an isosceles trapezoid have lengths of 48 - 7x and x + 8. What is the length of one of the legs?

Mathematics

1 answer:

sdas [7]3 years ago

4 0

Answer:

13 units

Step-by-step explanation:

The legs of an isosceles trapezoid are congruent, then

x + 8 = 48 - 7x ( add 7x to both sides )

8x + 8 = 48 ( subtract 8 from both sides )

8x = 40 ( divide both sides by 8 )

x = 5

Then

leg = x + 8 = 5 + 8 = 13

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39=1 3/10b what will be the answer show work

Lubov Fominskaja [6]

1 3/10·b=39
(1·10+3)·b/10=39
13b/10=39
13b=39·10
13b=390|:13
b=30

Answer: 30

4 0

2 years ago

Which is the best description for the graph below? A graph decreases from (0, 3) to (9, 0).

AURORKA [14]

Answer:

Step-by-step explanation:

As we move from (0, 3) to (9, 0), we see x increasing by 9. This is the 'run.'

At the same time, we see y decreasing by 3. This is the 'rise.'

The slope of this line (graph) is m = rise / run = -3/9, or m = -1/3

Since we witness a decrease in y as we move from the first point to the second, we immediately suspect that the slope is negative.

8 0

3 years ago

Mark invests $6,700 in an online savings account which gives 4.2% simple annual interest. He also invests $6,000 in a savings ac

dalvyx [7]

Answer:

$4,221.00 for online and 6,210.00 for the instore

Step-by-step explanation:

Calculation:

First, converting R percent to r a decimal

r = R/100 = 4.2%/100 = 0.042 per year.

Solving our equation:

A = 6700(1 + (0.042 × 15)) = 10921

A = $10,921.00

The total amount accrued, principal plus interest, from simple interest on a principal of $6,700.00 at a rate of 4.2% per year for 15 years is $10,921.00.

Calculation:

First, converting R percent to r a decimal

r = R/100 = 6.9%/100 = 0.069 per year.

Solving our equation:

A = 6000(1 + (0.069 × 15)) = 12210

A = $12,210.00

The total amount accrued, principal plus interest, from simple interest on a principal of $6,000.00 at a rate of 6.9% per year for 15 years is $12,210.00.

5 0

3 years ago

The function f(t) = 20 sin (pi over 5t) + 12 models the temperature of a periodic chemical reaction where t represents time in h

gregori [183]

The answer for the exercise shown above is the second option, which is: Maximum: 32°; minimum: −8°; period: 10 hours.
The explanation is shown below:
 You can make a graph of the function given in the problem above: f(t)=20Sin(π/5t)+12.
As you can see in the graph, the maximum point is at 32 over the y-axis, and the minimum is at -8.
The lenght of the repeating pattern of the function (Its period) is 10.

6 0

3 years ago

Express the given quantity as a single logarithm and simplify: 3logx+2log(y-2)-5logx

storchak [24]

Simplify each term.
Simplify 3log(x) by moving 3 inside the logarithm.
log(x^3)+2log(y−1)−5log(x)

Simplify 2log(y−1) by moving 2 inside the logarithm.
log(x^3)+log((y−1)^2)−5log(x)

Rewrite (y−1)^2 as (y−1)(y−1).
log(x^3)+log((y−1)(y−1))−5log(x)

Expand (y−1)(y−1) using the FOIL Method.
log(x^3)+log(y(y)+y(−1)−1(y)−1(−1))−5log(x)

Simplify each term.
log(x^3)+log(y^2−2y+1)+log(x^−5)

Remove the negative exponent by rewriting x^−5 as 1/x^5.
log(x^3)+log(y^2−2y+1)+log(1/x^5)

Combine logs to get log(x^3(y^2−2y+1))
log(x^3(y^2−2y+1))+log(1/x^5)

Combine logs to get log(x^3(y^2−2y+1)/x^5)
log(x^3(y^2−2y+1)/x^5)

Cancel x^3 in the numerator and denominator.
log(y^2−2y+1/x^2)

Rewrite 1 as 1^2.
y^2−2y+1^2/x^2

Factor by perfect square rule.
(y−1)^2/x^2

Replace into larger expression.

log((y−1)^2/x^2)

3 0

3 years ago