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2 years ago

21q - 7p and (3q - p)(7)

Mathematics

1 answer:

S_A_V [24]2 years ago

4 0

Answer:

Step-by-step explanation:

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Harper knows he is 50 yards from school. The map on his phone shows that the school is 2/3 inch from his current location. How f

lorasvet [3.4K]

Actual distance of the Harper from the school = 50 yards.

According to map the distance from the school = 34 inch.

We could say 34 inches represents 50 yards.

So, 1 inches could be represented by 50/34 = 25/17 yards.

In order to find the actual length of 3 inches in yards, we need to multiply 1 inches value of yards by 3.

Therefore, 3 inches = 3* 25/17 = 75/17 = 4 7/17 yeards in decimals 4.412 yards approximately.

Therefore, Harper from home is 4 7/17 yeards or 4.412 yards approximately if the map shows the distance as 3 inches.

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2 years ago

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Beth bought two pairs of earrings for $12.00 each. She also had to pay 6% sales tax. What was her total bill?

Darya [45]

I’m pretty sure it would be $12.72!

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2 years ago

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Quiz

Harman [31]

Answer:

h=-6 k=-30

Step-by-step explanation:

$y=x^2+12x+6$

$\mathrm{The\:vertex\:of\:an\:up-down\:facing\:parabola\:of\:the\:form}\:y=ax^2+bx+c\:\mathrm{is}\:x_v=-\frac{b}{2a}$

$a=1,\:b=12,\:c=6$

$x_v=-\frac{b}{2a}$

$x_v=-\frac{12}{2\cdot \:1}\\$

$x_v=-6\\$

$\mathrm{Plug\:in}\:\:x_v=-6\:\mathrm{to\:find\:the}\:y_v\:\mathrm{value}$

$y_v=\left(-6\right)^2+12\left(-6\right)+6\\y_v=-30\\\left(-6,\:-30\right)$

3 0

2 years ago

Plantea una ecuacion

Anna35 [415]

Answer:78phy

gigh[ood9..8d55.5.5.5.5.5.5.5.6.s7s47.4

Step-by-step explanation:

8 0

2 years ago

If f(x)=x^3-12x^2+35x-24f(x)=x 3 −12x 2 +35x−24 and f(8)=0f(8)=0, then find all of the zeros of f(x)f(x) algebraically.

Neporo4naja [7]

Answer:

The zeros of f(x) are: (x - 1), (x - 3) and (x - 8)

Step-by-step explanation:

Given

$f(x)=x^3-12x^2+35x-24$

$f(8) = 0$

Required

Find all zeros of the f(x)

If $f(8) = 0$ then:

$x = 8$

And $x - 8$ is a factor

Divide f(x) by x - 8

$\frac{f(x)}{x - 8} = \frac{x^3-12x^2+35x-24}{x - 8}$

Expand the numerator

$\frac{f(x)}{x - 8} = \frac{x^3 - 4x^2 -8x^2 + 3x + 32x - 24}{x - 8}$

Rewrite as:

$\frac{f(x)}{x - 8} = \frac{x^3 - 4x^2 + 3x - 8x^2 +32x - 24}{x - 8}$

Factorize

$\frac{f(x)}{x - 8} = \frac{(x^2 - 4x + 3)(x - 8)}{x - 8}$

Expand

$\frac{f(x)}{x - 8} = \frac{(x^2 -x - 3x + 3)(x - 8)}{x - 8}$

Factorize

$\frac{f(x)}{x - 8} = \frac{(x - 1)(x - 3)(x - 8)}{x - 8}$

$\frac{f(x)}{x - 8} = (x - 1)(x - 3)$

Multiply both sides by x - 8

$f(x) = (x - 1)(x - 3)(x - 8)$

<em>Hence, the zeros of f(x) are: (x - 1), (x - 3) and (x - 8)</em>

7 0

2 years ago