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3 years ago

Identify terms, like terms, coefficients, and constants 4y + 5 + 3y

Mathematics

1 answer:

Viefleur [7K]3 years ago

5 0

Answer: 7y+5

Step-by-step explanation:

Combine Like Terms:

4y+5+3y

(4y+3y)+(5)

7y+5

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Una piscina rectangular de 15 metros de largo por 9 metros de ancho està rodeada por un

stira [4]

Answer: 2 meters.

Step-by-step explanation:

Let w = width of the cement path.

Dimensions of pool : Length = 15 meters , width = 9 meters

Area of pool = length x width = 15 x 9 = 135 square meters

Along width cement path, the length of region = $2w+15$

width = $2w+9$

Area of road with pool = $(2w+15) (2w+9)$

$= 4w^2+30w+18w+135\\\\=4w^2+48w+135$

Area of road = (Area of road with pool ) -(area of pool)

$\Rightarrow\ 112 =4w^2+48w+135- 135\\\\\Rightarrow\ 112= 4w^2+48w\\\\\Rightarrow\ 4 w^2+48w-112=0\\\\\Rightarrow\ w^2+12w-28=0\ \ \ [\text{Divide both sides by 4}]\\\\\Rightarrow\ w^2+14w-2w-28=0\\\\\Rightarrow\ w(w+14)-2(w+14)=0\\\\\Rightarrow\ (w+14)(w-2)=0\\\\\Rightarrow\ w=-14\ or \ w=2$

width cannot be negative, so w=2 meters

Hence, the width of the road = 2 meters.

8 0

4 years ago

Summer Dave builds a crate to hold vegetables the volume of the crate is 48 ft.³ the base area of the crate is 16 ft.² what is t

MariettaO [177]

The height of crate is 3 feet

<em><u>Solution:</u></em>

Given that Summer Dave builds a crate to hold vegetables

Volume of crate = 48 cubic feet

Base area of crate = 16 square feet

To find: height of crate

The crate is of shape cuboid

The volume of cuboid is given as:

$volume = length \times width \times height$

We know that,

$Area\ of\ base = length \times width$

Therefore, volume becomes,

$volume = base\ area \times height$

Substituting the given values we get,

$48 = 16 \times h\\\\h = \frac{48}{16}\\\\h = 3$

Thus height of crate is 3 feet

6 0

3 years ago

Calculus. Find the area.

matrenka [14]

Answer:

8/3 square units.

Step-by-step explanation:

First, visualize the area. You can refer to the attachment below.

To find the area then, we will integrate y from x = 0 to x = 2. Therefore:

$\displaystyle A=\int_0^2x^2\, dx$

Integrate:

$\displaystyle A=\frac{1}{3}x^3\Big|_{0}^2$

Evaluate:

$\displaystyle A=\frac{1}{3}[2^3-0^3]=\frac{1}{3}(8)=\frac{8}{3}$

The area is 8/3 square units.

6 0

3 years ago

Read 2 more answers

Can someone please help me with my maths

Korvikt [17]

Answer:

Step-by-step explanation:

Recall that the "point-slope" form of a linear equation may be expressed as

y = mx + b,

where m is the gradient.

If m is negative, the gradient is negative.

If m is positive, the gradient is positive.

In our case, if we consider option A,

x + 3y = -2 (rearranging)

3y = -x -2

y = (-1/3) x - (2/3)

if we compare this to the general equation at the top, we can see that

gradient, = m = (-1/3) which is negative.

hence option a has a negative gradient.

3 0

3 years ago

Read 2 more answers

The height h (in feet) of an object dropped from a ledge after x seconds can be modeled by h(x)=−16x2+36 . The object is dropped

kakasveta [241]

Check the picture below.

$\bf ~~~~~~\textit{initial velocity in feet} \\\\ h(t) = -16t^2+v_ot+h_o \quad \begin{cases} v_o=\textit{initial velocity}&\\ \qquad \textit{of the object}\\ h_o=\textit{initial height}&\\ \qquad \textit{of the object}\\ h=\textit{object's height}&\\ \qquad \textit{at "t" seconds} \end{cases}$

so the object hits the ground when h(x) = 0, hmmm how long did it take to hit the ground the first time anyway?

$\bf h(x)=-16x^2+36\implies \stackrel{h(x)}{0}=-16x^2+36\implies 16x^2=36 \\\\\\ x^2=\cfrac{36}{16}\implies x^2 = \cfrac{9}{4}\implies x=\sqrt{\cfrac{9}{4}}\implies x=\cfrac{\sqrt{9}}{\sqrt{4}}\implies x = \cfrac{3}{2}~~\textit{seconds}$

now, we know the 2nd time around it hit the ground, h(x) = 0, but it took less time, it took 0.5 or 1/2 second less, well, the first time it took 3/2, if we subtract 1/2 from it, we get 3/2 - 1/2 = 2/2 = 1, so it took only 1 second this time then, meaning x = 1.

$\bf ~~~~~~\textit{initial velocity in feet} \\\\ h(x) = -16x^2+v_ox+h_o \quad \begin{cases} v_o=\textit{initial velocity}&0\\ \qquad \textit{of the object}\\ h_o=\textit{initial height}&\\ \qquad \textit{of the object}\\ h=\textit{object's height}&0\\ \qquad \textit{at "t" seconds}\\ x=\textit{seconds}&1 \end{cases} \\\\\\ 0=-16(1)^2+0x+h_o\implies 0=-16+h_o\implies 16=h_o \\\\[-0.35em] ~\dotfill\\\\ ~\hfill h(x) = -16x^2+16~\hfill$

quick info:

in case you're wondering what's that pesky -16x² doing there, is gravity's pull in ft/s².

4 0

3 years ago