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3 years ago

Sid Upp is four years younger than his brother Tip. Let x stand for Tip's age . Write an equation stating that Sid's age is 32.

Mathematics

1 answer:

vlada-n [284]3 years ago

4 0

Answer:

x+4=36

Step-by-step explanation:

you would subtract 4 on each side

x+4=36

-4 -4

x=32

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The area of the triangle below is 8.91 square inches. What is the length of the base?

Studentka2010 [4]

Answer:

8.1 inches

Step-by-step explanation:

Area of a triangle = ½ × base × height = Area

Rearranged =

$\frac{area}{ \frac{1}{2} \times height}$

Base is the length by the way.

$\frac{8.91}{ \frac{1}{2} \times 2.2 }$

=8.1 inches

6 0

3 years ago

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Ivan

Answer:

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Step-by-step explanation:

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1 year ago

Write an equation to match the fraction model shown in the reference image and solve the problem.

Pani-rosa [81]

Answer:

8/100 + 8/100 + 32/100 = 48/100

48/100 / 4/4 = 12/25

Step-by-step explanation:

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2 years ago

Determine the graph of the sinusoid with amplitude of 4 and period of 2π/3

DiKsa [7]

Answer:

Option b.

y = 4 sin (3x)

Step-by-step explanation:

To quickly solve this problem, we can use a graphing tool or a calculator to plot the equation.

Please see the attached image below, to find more information about the graph

The period is

T = 2π/3

This means the frequency in radians is

W = 2π/T = 3

The equation is either:

y = 4 cos(3x)

y = 4 sin (3x)

The correct answer is

Option b.

y = 4 sin (3x)

3 0

3 years ago

One of the roots of the quadratic equation dx^2+cx+p=0 is twice the other, find the relationship between d, c and p

scZoUnD [109]

Answer:

$c^2 = 9dp$

Step-by-step explanation:

Given

$dx^2 + cx + p = 0$

Let the roots be $\alpha$ and $\beta$

So:

$\alpha = 2\beta$

Required

Determine the relationship between d, c and p

$dx^2 + cx + p = 0$

Divide through by d

$\frac{dx^2}{d} + \frac{cx}{d} + \frac{p}{d} = 0$

$x^2 + \frac{c}{d}x + \frac{p}{d} = 0$

A quadratic equation has the form:

$x^2 - (\alpha + \beta)x + \alpha \beta = 0$

So:

$x^2 - (2\beta+ \beta)x + \beta*\beta = 0$

$x^2 - (3\beta)x + \beta^2 = 0$

So, we have:

$\frac{c}{d} = -3\beta$ -- (1)

and

$\frac{p}{d} = \beta^2$ -- (2)

Make $\beta$ the subject in (1)

$\frac{c}{d} = -3\beta$

$\beta = -\frac{c}{3d}$

Substitute $\beta = -\frac{c}{3d}$ in (2)

$\frac{p}{d} = (-\frac{c}{3d})^2$

$\frac{p}{d} = \frac{c^2}{9d^2}$

Multiply both sides by d

$d * \frac{p}{d} = \frac{c^2}{9d^2}*d$

$p = \frac{c^2}{9d}$

Cross Multiply

$9dp = c^2$

$c^2 = 9dp$

Hence, the relationship between d, c and p is: $c^2 = 9dp$

8 0

3 years ago