Ask question

Ask question

All categories

3 years ago

6

Write an expression to represent the area of a rectangle if the length of a rectangle is three units more than its width

1 answer:

Vitek1552 [10]3 years ago

5 0

We can call the width “x”. Since the length is three units more than its width, we can call the length “x+3”.

Thus, the area is x(x+3), which simplifies to x^2+3x.

You might be interested in

Use slopes and y-intercepts to determine if the lines y=−x2+14 and 3x−4y=−1 are parallel.

Keith_Richards [23]

Answer: the lines are not parallel.

Explanation: if you isolate the x in 3x^2-4y=-1, the equation is y=3/4x^2+1/4

So, they are not parallel because the slope is not the same.

7 0

2 years ago

The given function k is a composition of two functions m and n so that k(x) = (m ∘ n)(x). k (x) = StartFraction 2 Over RootIndex

kotykmax [81]

Answer:

$\frac{2}{\sqrt[4]{x} }$

Step-by-step explanation:

6 0

3 years ago

15. Before oil is added to a certain road surface material, the other ingredients are mixed. The mixture contains

mojhsa [17]

Answer:

sand = 3450 pounds

3/4 inch stones =6750 pounds

1 inch stones =4800 pounds.

Step-by-step explanation:

The mixture composition is :

45% three-quarter inch stones

32% inch stones and

sand for balance

The percent here is by mass of material used.

Weight of mixture is 15,000 pounds before oil addition.

Use the percent by weight for the two types of stones and calculate real values in pounds

Formula to apply is:

Mass percent = (mass of a material/ total mass of mixture) * 100

For 3/4 inch stones it will be:

$0.45= \frac{x}{15000} *100\\\\0.45=\frac{x}{150} \\0.45*150=x\\x=6750$

The 3/4 inch stones weigh 6750 pounds

The 1 inch stones will be

$32/100 * 15000 =4800$

The 1 inch stones weigh = 4800 pounds

Total for the two items : 4800+6750 =11550 pounds

The sand used weighs : 15000 - 11550 = 3450 pounds

8 0

3 years ago

Plllzzz im new and i neeed help

ella [17]

Answer:

4082

Step-by-step explanation:

Given

The composite object

Required

The volume

The object is a mix of a cone and a hemisphere

Such that:

<u>Cone</u>

$r = 10cm$ ---- radius (r = 20/2)

$h = 19cm$

<u>Hemisphere</u>

$r=10cm$

The volume of the cone is:

$V_1 = \frac{1}{3}\pi r^2h$

$V_1 = \frac{1}{3}\pi * 10^2 * 19$

$V_1 = \frac{1900}{3}\pi$

The volume of the hemisphere is:

$V_2 = \frac{2}{3}\pi r^3$

$V_2 = \frac{2}{3}\pi 10^3$

$V_2 = \frac{2000}{3}\pi$

So, the volume of the object is:

$V = V_1 + V_2$

$V = \frac{1900}{3}\pi + \frac{2000}{3}\pi$

$V = \frac{3900}{3}\pi$

$V = 1300\pi$

$V = 1300 * 3.14$

$V = 4082$

8 0

3 years ago

-18 > |-1 (3-2x)| > 18

12345 [234]

Answer:

No Solution

Step-by-step explanation:

Since -18 > 18 is not true, this inequality is always false

4 0

3 years ago

Read 2 more answers