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3 years ago

11

help!! Katherine O’Donnell obtain a personal loan of $3000 at 14% for 24 months. The monthly payment is $144. What is the intere

st of the first payment?

1 answer:

lina2011 [118]3 years ago

8 0

3000*0.14*24=??? Then uuu add the 144 to your answer to get your answer

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A rabbit can run short distances at a rate of 35 miles per hour. A fox can run short distances at a rate of 21 miles per half ho

cricket20 [7]

Answer:

the fox is faster by 0.12.

Step-by-step explanation:

7 0

3 years ago

Question on the image

kodGreya [7K]

Let $x,y$ be the legs of the triangle, with $x[tex]\mathrm{area}_{\rm square} = y^2$

$\mathrm{area}_{\rm triangle} = \dfrac12 xy$

The square has 3 times the area of the triangle, so

$y^2 = \dfrac32 xy$

Meanwhile, in the triangle we have

$\tan(\theta) = \dfrac xy$

Now,

$y^2 = \dfrac32 xy \implies \dfrac23 = \dfrac xy \implies \boxed{\tan(\theta) = \dfrac23}$

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2 years ago

If X- 6 is the only x-intercept of the graph of a quadratic equation, which statement best describes the discriminant of the equ

Verdich [7]

The rule, letting d=discriminant:

d>0, two real solutions

d=0, one real solution

d<0, no real solution (but there are two imaginary solutions)

So if it has only one real solution, the discriminant is equal to zero.

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3 years ago

Read 2 more answers

For question 12, use the graph to guess one zero for the polynomial whose graph is shown. Then, factor completely.

SVEN [57.7K]

Answer:

x³ - 6·x² + 6·x + 8 = (x - 4) × (x - (1 + √3)) × (x + 1 + √3)

Step-by-step explanation:

From the given graph of the polynomial, x³ - 6·x² + 6·x + 8, we have that one of the zeros of the polynomial, which is one point the graph cuts the x-axis is given as follows;

At a zero, we have, x = 4

∴ x - 4 = 0 is a factor of the polynomial

Dividing the given polynomial by (x - 4), gives;

<u>x² - 2·x - 2</u>

(x³ - 6·x² + 6·x + 8)÷(x - 4)

<u>x³ -4·x²</u>

-2·x² + 6·x + 8

<u>-2·x² + 8·x</u>

-2·x + 8

<u>-2·x + 8</u>

<u>0</u>

∴ (x³ - 6·x² + 6·x + 8)÷(x - 4) = x² - 2·x - 2

∴ (x - 4) × (x² - 2·x - 2) = x³ - 6·x² + 6·x + 8

Factorizing x² - 2·x - 2, gives;

x = (2 ± √((-2)² - 4×1×(-2)))/(2 × 1) = (2 ± 2·√(3)/(2) = (1 ± √(3)

Therefore, x = 1 + √3, (x - (1 + √3)) = 0 or x = 1 - √3, (x - (1 - √3)) = 0 which gives the factors, after factorizing completely as follows;

x³ - 6·x² + 6·x + 8 = (x - 4) × (x - (1 + √3)) × (x + 1 + √3)

4 0

3 years ago

During 712 months of hibernation, a black bear experienced a weight loss of 64.4 pounds. On average, what was the bear's weight

Romashka-Z-Leto [24]

Total pounds of weight loss of bear in $7 \frac{1}{2}$ months = 64.4 pounds

We have to determine the bear's weight change per month.

We will use unitary method to solve this problem.

So, total pounds of weight loss in 1 month = $64.4 \div 7 \frac{1}{2}$

= $64.4 \div \frac{15}{2}$

= $\frac{64.4 \times 2}{15}$

= 8.58

Since, we have to round to the nearest tenth.

Therefore, the bear's weight change per month is 8.6 pounds.

6 0

4 years ago