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3 years ago

12

Guys, please help out.

1 answer:

Margarita [4]3 years ago

5 0

Answer:

5x + 13

Step-by-step explanation:

The 2 binomials in algebraic form are

2x + 8 and 3x + 5

Summing them

2x + 8 + 3x + 5 ← collect like terms

(2x + 3x) + (8 + 5)

= 5x + 13

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To change knots per hour, use the expression 1.15k, where k is the speed in in knots per hour. A plane is flying at 300 knots pe

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1.15(300).
Substitute k for 300, since the plane is flying 300 knots per hour, and k=knots per hour.

1.15(300)=345 miles per hour
After multiplying, 300 by 1.15, you get 345 miles per hour. This is the speed of the plane from Knots to MPH (Miles Per Hour).

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3 years ago

Read 2 more answers

E. Find the slope of the line that passes through the points (1, 3) and (9,7),

Zolol [24]

Answer:

0.5

Step-by-step explanation:

→ Find the difference in y

7 - 3 = 4

→ Find the difference in x

9 - 1 = 8

→ Divide the results

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3 years ago

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HELP PLEASE! ONLY 5 MORE MINUTES! Look at the following shape below: Please find the PERIMETER of the following shape using the

Evgen [1.6K]

Answer:

The perimeter is $P=73.12\ cm$

Step-by-step explanation:

we know that

The perimeter of the figure is equal to the circumference of a semicircle

plus the perimeter of a square minus the diameter of the circle

so

$P=\pi r+4(b)-D$

we have

$r=8\ cm$

$b=16\ cm$

$D=16\ cm$

The diameter of the circle is equal to the length side of the square

substitute

$P=(3.14)(8)+4(16)-16$

$P=25.12+64-16$

$P=73.12\ cm$

7 0

3 years ago

There are 24 basketball teams competing in a tournament. After each round half the teams are eliminated. This situation can be m

BaLLatris [955]

Answer:

b(x)=24(1/2)x

Step-by-step explanation:

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3 years ago

In the previous part, we obtained dy dx = 3t2 − 27 −2t . Next, find the points where the tangent to the curve is horizontal. (En

mina [271]

Answer:

(27.55, 7.22), (-11.3, 3.21).

Step-by-step explanation:

When is the tangent to the curve horizontal?

The tangent curve is horizontal when the derivative is zero.

The derivative is:

$\frac{dy}{dx} = 3t^{2} - 2t - 27$

Solving a quadratic equation:

Given a second order polynomial expressed by the following equation:

$ax^{2} + bx + c, a\neq0$ .

This polynomial has roots $x_{1}, x_{2}$ such that $ax^{2} + bx + c = a(x - x_{1})*(x - x_{2})$ , given by the following formulas:

$x_{1} = \frac{-b + \sqrt{\bigtriangleup}}{2*a}$

$x_{2} = \frac{-b - \sqrt{\bigtriangleup}}{2*a}$

$\bigtriangleup = b^{2} - 4ac$

In this question:

$3t^{2} - 2t - 27 = 0$

So

$a = 3, b = -2, c = -27$

Then

$\bigtriangleup = b^{2} - 4ac = (-2)^{2} - 4*3*(-27) = 328$

So

$t_{1} = \frac{-(-2) + \sqrt{328}}{2*3} = 3.35$

$t_{2} = \frac{-(-2) - \sqrt{328}}{2*3} = -2.685$

Enter your answers as a comma-separated list of ordered pairs.

We found values of t, now we have to replace in the equations for x and y.

t = 3.35

$x = t^{3} - 3t = (3.35)^{3} - 3*3.35 = 27.55$

$y = t^{2} - 4 = (3.35)^2 - 4 = 7.22$

The first point is (27.55, 7.22)

t = -2.685

$x = t^{3} - 3t = (-2.685)^3 - 3*(-2.685) = -11.3$

$y = t^{2} - 4 = (-2.685)^2 - 4 = 3.21$

The second point is (-11.3, 3.21).

8 0

3 years ago