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2 years ago

4

James drove from Baltimore to Washington DC. The distance is 45 miles and the trip took 0.75 of an hour. How fast was James driv

ing?

1 answer:

love history [14]2 years ago

4 0

Answer:

60mph

Step-by-step explanation:

60mph means 1 mile per minute and .75 of an hour(60min) is 45 minute.

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CD is a perpendicular bisector of . Which of the following could be the slopes of the two lines if plotted on a coordinate grid?

NISA [10]

Step-by-step explanation:

A is the correct answer of this question

8 0

2 years ago

Let $f(x) = x^{10}-8x^8-8x^3+12x^2-5x-5$. Without using long division (which would be horribly nasty!), find the remainder when

valentina_108 [34]

Solution:

$f(x) = x^{10}-8x^8-8x^3+12x^2-5x-5$

We have to find the remainder when f(x) is divided by $x^2-1.$

x²-1=0

x= $\pm1$

$f(1)=1^{10}-8(1)^8-8(1)^3+12(1)^2-5(1)-5=1-8-8+12-5-5=1-16+12-10=13-26=-13\\\\f(-1)=(-1)^{10}-8(-1)^8-8(-1)^3+12(-1)^2-5(-1)-5=1-8+8+12+5-5=1+12=13$

So, remainder is 13 and -13.

5 0

2 years ago

Read 2 more answers

which of the following is equivalent to 3 sqrt 32x^3y^6 / 3 sqrt 2x^9y^2 where x is greater than or equal to 0 and y is greater

Nutka1998 [239]

Answer:

$\frac{\sqrt[3]{16y^4}}{x^2}$

Step-by-step explanation:

The options are missing; However, I'll simplify the given expression.

Given

$\frac{\sqrt[3]{32x^3y^6}}{\sqrt[3]{2x^9y^2} }$

Required

Write Equivalent Expression

To solve this expression, we'll make use of laws of indices throughout.

From laws of indices $\sqrt[n]{a} = a^{\frac{1}{n}}$

So,

$\frac{\sqrt[3]{32x^3y^6}}{\sqrt[3]{2x^9y^2} }$ gives

$\frac{(32x^3y^6)^{\frac{1}{3}}}{(2x^9y^2)^\frac{1}{3}}$

Also from laws of indices

$(ab)^n = a^nb^n$

So, the above expression can be further simplified to

$\frac{(32^\frac{1}{3}x^{3*\frac{1}{3}}y^{6*\frac{1}{3}})}{(2^\frac{1}{3}x^{9*\frac{1}{3}}y^{2*\frac{1}{3}})}$

Multiply the exponents gives

$\frac{(32^\frac{1}{3}x*y^{2})}{(2^\frac{1}{3}x^{3}*y^{\frac{2}{3}})}$

Substitute $2^5$ for 32

$\frac{(2^{5*\frac{1}{3}}x*y^{2})}{(2^\frac{1}{3}x^{3}*y^{\frac{2}{3}})}$

$\frac{(2^{\frac{5}{3}}x*y^{2})}{(2^\frac{1}{3}x^{3}*y^{\frac{2}{3}})}$

From laws of indices

$\frac{a^m}{a^n} = a^{m-n}$

This law can be applied to the expression above;

$\frac{(2^{\frac{5}{3}}x*y^{2})}{(2^\frac{1}{3}x^{3}*y^{\frac{2}{3}})}$ becomes

$2^{\frac{5}{3}-\frac{1}{3}}x^{1-3}*y^{2-\frac{2}{3}}$

Solve exponents

$2^{\frac{5-1}{3}}*x^{-2}*y^{\frac{6-2}{3}}$

$2^{\frac{4}{3}}*x^{-2}*y^{\frac{4}{3}}$

From laws of indices,

$a^{-n} = \frac{1}{a^n}$ ; So,

$2^{\frac{4}{3}}*x^{-2}*y^{\frac{4}{3}}$ gives

$\frac{2^{\frac{4}{3}}*y^{\frac{4}{3}}}{x^2}$

The expression at the numerator can be combined to give

$\frac{(2y)^{\frac{4}{3}}}{x^2}$

Lastly, From laws of indices,

$a^{\frac{m}{n} = \sqrt[n]{a^m}$ ; So,

$\frac{(2y)^{\frac{4}{3}}}{x^2}$ becomes

$\frac{\sqrt[3]{(2y)}^{4}}{x^2}$

$\frac{\sqrt[3]{16y^4}}{x^2}$

Hence,

$\frac{\sqrt[3]{32x^3y^6}}{\sqrt[3]{2x^9y^2} }$ is equivalent to $\frac{\sqrt[3]{16y^4}}{x^2}$

8 0

2 years ago

Adiya said that the first step to solving the quadratic equation x2 + 6 = 20x by completing the square was to divide 6 by 2, squ

kicyunya [14]

Adiya’s method is not correct. To form a perfect square trinomial, the constant must be isolated on one side of the equation. Also, the coefficient of the term with an exponent of 1 on the variable is used to find the constant in the perfect square trinomial. Adiya should first get the 20x term on the same side of the equation as x2. Then she would divide 20 by 2, square it, and add 100 to both sides.

6 0

2 years ago

Read 2 more answers

The perimeter of a rectangle is represented by 4x2 + 5x − 2. The perimeter of a smaller rectangle is represented by x2 + 3x + 5.

REY [17]

<span>4x2 +5x - 2 -x2 +3x +5 Simplified: 3x2 + 8x +3 In order to determine the difference in size, you must subtract the perimeter of the smaller rectangle from the perimeter of the larger rectangle.</span>

6 0

2 years ago

Read 2 more answers