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3 years ago

Which of the following polynomials is in standard form?

Mathematics

2 answers:

Tju [1.3M]3 years ago

6 0

Answer:

The 2nd one

Step-by-step explanation:

Allushta [10]3 years ago

5 0

Answer:

running a two year contract is the best i have to do with my life in this world and it has a great way to go through the

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Which statement is the inverse of the following? If the sum of the measures of two angles is 90°, then the angles are complement

lara [203]

A. if two angles are complementary angles, then the sum of the measures of two angles is 90°.

3 0

3 years ago

Read 2 more answers

A new car is purchased for $17,000 and over time its value depreciates by one half every 3 years. What is the value of the car 1

Jet001 [13]

Answer:

The answer is "$238".

Step-by-step explanation:

Current worth $= \$ 17,000$

depreciates by $\frac{1}{2}$ in 3 years.

time= 19 years

depreciates rate=?

Using formula:

$\to \text{Worth= Current worth}(1- \frac{\text{depreciates rate}}{100})^{time}$

$\to A_t=A_0(1-\frac{r}{100})^t$

calculates depreciate value in 3 year $= \frac{1}{2} \times 17,000$

$= 8,500$

so,

$A_t=8,500\\\\A_0=17,000\\\\t=3\ years$

$\to A_t=A_0(1-\frac{r}{100})^t\\\\\to 8,500= 17,000(1-\frac{r}{100})^3\\\\\to \frac{8,500}{17,000}= (1-\frac{r}{100})^3\\\\\to \frac{1}{2}= (1-\frac{r}{100})^3\\\\\to (\frac{1}{2})^{\frac{1}{3}}= (1-\frac{r}{100})\\\\\to 0.793700526 = (1-\frac{r}{100})\\\\\to \frac{r}{100} = (1-0.793700526)\\\\\to \frac{r}{100} = (1-0.8)\\\\\to r= 0.2 \times 100 \\\\\to r= 20 \%$

depreciates rate= 20%

$\to \text{Worth= Current worth}(1- \frac{\text{depreciates rate}}{100})^{time}$

$= \$ 17,000 (1- \frac{20}{100})^{19}\\\\= \$ 17,000 (1-0.2)^{19}\\\\= \$ 17,000 (0.8)^{19}\\\\= \$ 17,000 \times 0.014\\\\= \$ 238$

4 0

3 years ago

Read 2 more answers

The parallelogram shown below has an area of <br> 20 units squared. Find the missing base

valkas [14]

Answer: 5 units

Step-by-step explanation:

The area of a parallelogram is $A=bh$ . SInce we know the area and the height, we can directly plug the values into this equation and solve for the base.