Ask question

Ask question

All categories

3 years ago

5

70% of 80 is what number

2 answers:

Nat2105 [25]3 years ago

6 0

First, you would change 70% into a decimal, 0.70. Then, you would multiply 80 by 0.70, and that is the answer. 70% of 80 is 56.

rosijanka [135]3 years ago

3 0

It would be 56,
I wish you luck.

You might be interested in

Andy's average driving speed for a 4 hour trip was 48 miles per hour. During the first 3 hours he drove 50 miles per hour. What

seraphim [82]

Answer:

Step-by-step explanation:

Distance = 48 * 4

Distance = 192

~~~~~~~~~~~~~~~

3*50 = 150

192 - 150 = 42

42 MPH for the last hour

3 0

3 years ago

Simplify: (Algebra)

kodGreya [7K]

Answer:

$2j-3o-5e+l-j+4o+6e= e+j+l+o$

$-7d+a+2v-i+8d-v+2i+d^2 = d^2+a+d+i+v$

$4a+12r-i+a^2-11r-3a+n+a^3 = a^3+a^2+a-i+n+r$

$r^3-11a+12a-sh+2sh+i+d= r^3+a+sh+i+d$

$t^5+a-2a+2r-r+l-2l+a^4+3n-2n = t^5-a+r-l+a^4+n$

$-11i^2-r+2r+a^3-4d+3d-a = -11i^2+r+a^3-d-a$

$-9r+10r+a-s^3-2a+2s^3+u-l = r-a+s^3+u-l$

$-4d+2a^3-a^3+5n+iz-4n = -4d+a^3+n+iz$

$16a^5-2(l-y+y3)-a+d^2-d^3+in = 16a^5-2l-2y+2y3-a+d^2-d^3+in$

$-5z-e+2e-y^5-n+2n-a^5-l+l^3-i^3 = -5z+e-y^5+n-a^5-l+l^3-i^3$

$-2m+a-m^3-2m^5-a^3-d+l+i^4= -2m+a-m^3-2m^5-a^3-d+l+i^4$

$-4t^3+5t^3+3o-2o+2m-m-r^3-is+2is = t^3+o+m-r^3+is$

$-5j+h-2h+4j-3a+6a-d^3 = -5j-h+4j+3a-d^3$

$r^3-3u+2u-st+a^3-2a^3-5m+4m = r^3-u-st-a^3-m$

$-6b^5-a+4h-3h+18a^2+3a-d^4+i^2-2i^2-4r^4= -6b^5+3a-a+4h-3h+18a^2-d^4+i^2-2i^2-4r^4$

$13s^5-2o+5n^3-12s^5-6n^3+3o+a^5=s^5+o-n^3+a^5$

$-17m+16m-3u+r^3+2u-2r^3+a^4+d^4= -1m+-u-r^3+a^4+d^4$

Step-by-step explanation:

$a.\ 2j-3o-5e+l-j+4o+6e$

Collect Like Terms

$-5e+6e+2j-j+l+4o-3o$

$e+j+l+o$

Hence:

$2j-3o-5e+l-j+4o+6e= e+j+l+o$

$b.\ -7d+a+2v-i+8d-v+2i+d^2$

Collect Like Terms

$d^2+a+8d-7d-i+2i+2v-v$

$d^2+a+d+i+v$

Hence:

$-7d+a+2v-i+8d-v+2i+d^2 = d^2+a+d+i+v$

$c.\ 4a+12r-i+a^2-11r-3a+n+a^3$

Collect Like Terms

$a^3+a^2+4a-3a-i+n+12r-11r$

$a^3+a^2+a-i+n+r$

Hence:

$4a+12r-i+a^2-11r-3a+n+a^3 = a^3+a^2+a-i+n+r$

$d.\ r^3-11a+12a-sh+2sh+i+d$

Evaluate like terms

$r^3+a+sh+i+d$

Hence:

$r^3-11a+12a-sh+2sh+i+d= r^3+a+sh+i+d$

$e.\ t^5+a-2a+2r-r+l-2l+a^4+3n-2n$

Evaluate like terms

$t^5-a+r-l+a^4+n$

Hence:

$t^5+a-2a+2r-r+l-2l+a^4+3n-2n = t^5-a+r-l+a^4+n$

$f.\ -11i^2-r+2r+a^3-4d+3d-a$

Evaluate Like Terms

$-11i^2+r+a^3-d-a$

Hence:

$-11i^2-r+2r+a^3-4d+3d-a = -11i^2+r+a^3-d-a$

$g.\ -9r+10r+a-s^3-2a+2s^3+u-l$

Collect Like Terms

$-9r+10r+a-2a+2s^3-s^3+u-l$

$r-a+s^3+u-l$

Hence:

$-9r+10r+a-s^3-2a+2s^3+u-l = r-a+s^3+u-l$

$h.\ -4d+2a^3-a^3+5n+iz-4n$

Collect Like Terms

$-4d+2a^3-a^3+5n-4n+iz$

$-4d+a^3+n+iz$

Hence:

$-4d+2a^3-a^3+5n+iz-4n = -4d+a^3+n+iz$

$i.\ 16a^5-2(l-y+y3)-a+d^2-d^3+in$

Open bracket

$16a^5-2l-2y+2y3-a+d^2-d^3+in$

Hence:

$16a^5-2(l-y+y3)-a+d^2-d^3+in = 16a^5-2l-2y+2y3-a+d^2-d^3+in$

$j.\ -5z-e+2e-y^5-n+2n-a^5-l+l^3-i^3$

Evaluate like terms

$-5z+e-y^5+n-a^5-l+l^3-i^3$

Hence:

$-5z-e+2e-y^5-n+2n-a^5-l+l^3-i^3 = -5z+e-y^5+n-a^5-l+l^3-i^3$

$k.\ m-3m-a+2a-m^3-2m^5-a^3-d+l+i^4$

Evaluate like terms

$-2m+a-m^3-2m^5-a^3-d+l+i^4$

Hence:

$-2m+a-m^3-2m^5-a^3-d+l+i^4= -2m+a-m^3-2m^5-a^3-d+l+i^4$

$l.\ -4t^3+5t^3+3o-2o+2m-m-r^3-is+2is$

Evaluate like terms

$t^3+o+m-r^3+is$

Hence

$-4t^3+5t^3+3o-2o+2m-m-r^3-is+2is = t^3+o+m-r^3+is$

$m.\ -5j+h-2h+4j-3a+6a-d^3$

Evaluate like terms

$-5j-h+4j+3a-d^3$

Hence:

$-5j+h-2h+4j-3a+6a-d^3 = -5j-h+4j+3a-d^3$

$n.\ r^3-3u+2u-st+a^3-2a^3-5m+4m$

Evaluate like terms

$r^3-u-st-a^3-m$

Hence:

$r^3-3u+2u-st+a^3-2a^3-5m+4m = r^3-u-st-a^3-m$

$o.\ -6b^5-a+4h-3h+18a^2+3a-d^4+i^2-2i^2-4r^4$

Collect Like Terms

$-6b^5+3a-a+4h-3h+18a^2-d^4+i^2-2i^2-4r^4$

$-6b^5+2a+h+18a^2-d^4-i^2-4r^4$

Hence:

$-6b^5-a+4h-3h+18a^2+3a-d^4+i^2-2i^2-4r^4= -6b^5+3a-a+4h-3h+18a^2-d^4+i^2-2i^2-4r^4$

$p.\ 13s^5-2o+5n^3-12s^5-6n^3+3o+a^5$

Collect Like Terms

$13s^5-12s^5-2o+3o+5n^3-6n^3+a^5$

$s^5+o-n^3+a^5$

Hence:

$13s^5-2o+5n^3-12s^5-6n^3+3o+a^5=s^5+o-n^3+a^5$

$q.\ -17m+16m-3u+r^3+2u-2r^3+a^4+d^4$

Collect Like Terms

$-17m+16m+2u-3u+r^3-2r^3+a^4+d^4$

$-m+-u-r^3+a^4+d^4$

Hence:

$-17m+16m-3u+r^3+2u-2r^3+a^4+d^4= -1m+-u-r^3+a^4+d^4$

7 0

4 years ago

PLS PLS PLS PLS PLS PLS HELP MEEEEEEEEE

horrorfan [7]

The measure of the arc WT in the circle is 51 degrees.

<h3>How to find the measure of an arc when tangent intersect?</h3>

The measure of the angle formed by the intersection of two tangents outside a circle is half the positive difference of the measure of the intercepted arc.

Therefore,

m∠RPS = arc RS - arc WT / 2

m∠RPS = 35°

arc RS = 121°

arc WT = ?

35 = 121 - x / 2

cross multiply

70 = 121 - x

70 - 121 = -x

x = 51°

Therefore,

arc WT = 51°

learn more on tangent here: brainly.com/question/16038958

#SPJ1

7 0

2 years ago

Find the quotient-you must show your thinking to support your answer. <br> -8 divided by (-2/3)

adell [148]

When you divide fractions you need to flip.
-8/1 x -3/2
Negative times negative is positive
24/2
12 is your answer

8 0

3 years ago

Read 2 more answers

Which solution shown below contains an error?

iren2701 [21]

Answer:

its B

Step-by-step explanation:

trust me i got it right on edge

7 0

3 years ago

Read 2 more answers