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3 years ago

6

Combine like terms for 5b+7-3b

2 answers:

sammy [17]3 years ago

6 0

5b and -3b are like terms. Add -3b to 5b, leaving 2b
Answer is 2b + 7

zubka84 [21]3 years ago

6 0

1) 5b-3b= 2b
2) Just add 7 to the added term- 2b+7

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I need help doing this.

melisa1 [442]

Answer:

6.63

Step-by-step explanation:

6.63324958071

the hundredth spot is the first 3, so you go to the second 3, because its under 5, the second 3 stays the same and doesn't round up to 4.

3 0

3 years ago

Help please!!! Solve the inequality. Show your work.

Rashid [163]

| r + 3 | > 7

r + 3 > 7 r + 3 < -7
r > 7 - 3 r < -7 - 3
r > 4 r < -10

so, r > 4 and r < -10
==========================
what percent of 60 is 12
percent formula : is/of = %/100
percent = x
is = 12
of = 60
now sub
12/60 = x/100
cross multiply
60x = 12 * 100
60x = 1200
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7 0

2 years ago

Read 2 more answers

Simplify. Assume that all variables are positive.

Alja [10]

I'm confused... Should there be more pictures? This in incomplete...

3 0

2 years ago

What is the unit rate of 5/4 square yards per 1/3 hour??????

Citrus2011 [14]

(5/4 yd^2)/ (1/3 hour)= 15/4 yd^2/hour.

The unit rate is 3 3/4 yd^2/hour~

5 0

3 years ago

Find the points on the curve where the tangent is horizontal or vertical. If you have a graphing device, graph the curve to chec

klasskru [66]

Answer:

There is a horizontal tangent at (0,-4)

The tangent is vertical at (-2,-3) and (2,-3).

Step-by-step explanation:

The given function is defined parametrically by the equations:

$x=t^3-3t$

and

$y=t^2-4$

The tangent function is given by:

$\frac{dy}{dx}=\frac{\frac{dy}{dt} }{\frac{dx}{dt} }$

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

The tangent is vertical at when $\frac{dx}{dt}=0$

$\implies \frac{3t^2-3}{2t}=0$

$\implies 3t^2-3=0$

$\implies 3t^2=3$

$\implies t^2=1$

$\implies t=\pm1$

When t=1,

$x=1^3-3(1)=-2$ and $y=1^2-4=-3$

When t=-1,

$x=(-1)^3-3(-1)=2$ and $y=(-1)^2-4=-3$

The tangent is vertical at (-2,-3) and (2,-3).

The tangent is horizontal, when $\frac{dy}{dx}=0$ or $\frac{dy}{dt}=0$

$\implies 2t=0$

$\implies t=0$

When t=0,

$x=0^3-3(0)=0$ and $y=0^2-4=-4$

There is a horizontal tangent at (0,-4)

5 0

2 years ago