All categories

2 years ago

Simplify the expression. 4(-8x + 5) – (-33x – 26) = ______

Mathematics

1 answer:

ziro4ka [17]2 years ago

7 0

Answer:

14+33x

Step-by-step explanation:

you have to break it up

4x(-3)+33x+26 MILTIPLY

-12+33x+26 CALCULATE

Solution is 14+33x

I HOPE THIS HELPS YOU

You might be interested in

If the 10th of the 10 consecutive numbers is 10, what is their sum?

liq [111]

Represent these consecutive numbers (assuming that they are all integers):

x
x+1
x+2
x+3
x+4
x+5
and so on
x+8
x+9 is the tenth number. x+9 = 10, so x = 9.

Think of it this way: there are 10 consecutive numbers, and the last one is 10.

Working backwards, we get the sequence 10, 9, ... 3, 2, 1.

The sum of such an arith sequence is equal to the count of the numbers times the average of the first and last terms:

sum here = 10(1+10)/2 = 5(11) = 55 (answer)

7 0

2 years ago

Part 4. Verbs

Pie

Answer:

MORE STUFF

Step-by-step explanation:

AD STUFF TO STUFF AND YOU GET MORE STUFF

8 0

3 years ago

Sarah opened new savings account on Tuesday with a deposit of $70. She made a deposit of $30 on Friday and a withdrawal of $60 o

irakobra [83]

Answer:

x=70+30-60

Step-by-step explanation:

Word to my mother

3 0

1 year ago

If c is the curve given by \mathbf{r} \left( t \right = \left( 1 5 \sin t \right \mathbf{i} \left( 1 3 \sin^{2} t \right \mathbf

jonny [76]

With the curve $C$ parameterized by

$C:\mathbf r(t)=\underbrace{15\sin t}_{x(t)}\,\mathbf i+\underbrace{13\sin^2t}_{y(t)}\,\mathbf j+\underbrace{12\sin^3t}_{z(t)}\,\mathbf k$

with $0\le t\le\dfrac\pi2$ , and given the vector field

$\mathbf f(x,y,z)=x\,\mathbf i+y\,\mathbf j+z\,\mathbf k$

the work done by $\mathbf f$ on a particle moving on along $C$ is given by the line integral

$\displaystyle\int_C\mathbf f\cdot\mathrm d\mathbf r=\int\limits_{t=0}^{t=\pi/2}\mathbf f(x(t),y(t),z(t))\cdot\frac{\mathrm d\mathbf r(t)}{\mathrm dt}\,\mathrm dt$

where

$\mathrm d\mathbf r=(15\cos t\,\mathbf i+26\sin t\cos t\,\mathbf j+36\sin^2t\cos t\,\mathbf k)\,\mathrm dt$

The integral is then

$\displaystyle\int_0^{\pi/2}(15\sin t\,\mathbf i+13\sin^2t\,\mathbf j+12\sin^3t\,\mathbf k)\cdot(15\cos t\,\mathbf i+13\sin2t\,\mathbf j+18\sin t\sin2t\,\mathbf k)\,\mathrm dt$
$=\displaystyle\int_0^{\pi/2}(432\sin^5t\cos t+338\sin^3t\cos t+225\sin t\cos t$
$=269$

6 0

3 years ago

Please please help me with this

swat32

N=1→an=a1 (first term)=16 (on the graph for n=1)→First term = 16
n=2→an=a2 (second term) = 4 (on the graph for n=2)→Second term = 4

ratio=(Second term)/(First term)=a2/a1=4/16
Simplifying the fraction dividing the numerator (4) by 4 and the denominator (16) by 4:
ratio=(4/4)/(16/4)→ratio=1/4

Answer: Option A. First term = 16, ratio = 1/4

7 0

3 years ago