170 students or 85% of the students went to attend college what are the total number of students?

Answer:

2.29883

formula

[tex]c = 2 \times \pi \times r[/tex]

solving for (r) radius

[tex]r = \frac{c}{2\pi} = \frac{14.44}{2 \times \pi} [/tex]

= 2.29883

Answer:

26/35

Step-by-step explanation:

1. First to divide the 3 1/2 by 2 3/5 you have to turn them both into improper fractions

First take 3 1/2. You have to multiply the whole number (3) by the denominator (2) and you would get 6. Then you would add then you add the product (6) to the numerator (1) and get 7.

You keep the denominator the same so the improper fraction is 7/2

Do the same thing to 2 3/5 and the improper fraction is 13/5

2. Now we can divide 13/5 by 7/2 using "keep, change, flip"

Keep: 13/5

Change: division to multiplcation

Flip: 7/2 to make 2/7

Your new equation is 13/5 × 2/7. Multiplcation is easy so you just have to multiply staight across: 13 × 2 and 5 × 7 giving you 26/35

If you divide 35 by 26 you will get 1.34 and a bunch of other numbers but I usually stop at two decimal places

hope this helps :)

Answer:

$14.40 is the discount price.

Step-by-step explanation:

0.2 x 18 = 3.6

18 - 3.6 = 14.4

Answer:

5 Motorcycles are used

30÷6= 5

5×5= 25

30-24=5

so the answer is 5

Answer:

5 out of the 30 motorcycles are used

Step-by-step explanation:

First, we should look at what we know. We know that there are 30 motorcycles in total. As a fraction, that'd be 30/30. We also know that of that 30, the equivalent of 5/6 are brand new.

Our next step is to figure out how much 5/6 is out of 30. Since we know that in total there 30/30 motorcycles, we need to convert 5/6 to have a denominator of 30. So what times 6 equals 30.

30 ÷ 6 = ?    opposite of 6 x ? = 30

5                   divide

So now we now that 6 times 5 equals 30. What we do to the denominator we must do to the numerator.

[tex]\frac{5}{6}[/tex] × [tex]\frac{5}{5}[/tex]    

[tex]\frac{25}{30}[/tex]

Now we know that 25 of the 30 motorcycles are new and all we need to do is subtract that 25 from the total 30 to get 5. This tells us that 5 of the total motorcycles are used.

I hope this helped :)

Answer:

I think that you should take your time and answer, every one can help you

Step-by-step explanation:

Step-by-step explanation:

ok give question and intro new sis

Answer:

the correct answer is B

Step-by-step explanation:

I got it correct on 2021 edge

Answer:

graph X only

Step-by-step explanation:

because with the rate of change it makes a straight line

Step-by-step explanation:

[tex] = {( \frac{27}{8} )}^{ \frac{1}{3} } \times ( \frac{243}{32} )^{ \frac{1}{5} } \div {( \frac{2}{3} )}^{2} [/tex]

[tex] = { ({ (\frac{3}{2} )}^{3}) }^{ \frac{1}{3} } \times {( {( \frac{3}{2}) }^{5} )}^{ \frac{1}{5} } \div {( \frac{2}{3} )}^{2} [/tex]

[tex] = {( \frac{3}{2} )}^{3 \times \frac{1}{3} } \times {( \frac{3}{2} )}^{5 \times \frac{1}{5} } \times {( \frac{3}{2} )}^{2} [/tex]

[tex] = \frac{3}{2} \times \frac{3}{2} \times {( \frac{3}{2} )}^{2} [/tex]

[tex] = {( \frac{3}{2} )}^{1 + 1 + 2} [/tex]

[tex] = {( \frac{3}{2} )}^{4} \: or \: \frac{81}{16} [/tex]

[tex]\large\underline{\sf{Solution-}}[/tex]

[tex]\sf{\longmapsto{\bigg( \dfrac{27}{8} \bigg)^{\frac{1}{3}} \times \Bigg[\bigg( \dfrac{243}{32} \bigg)^{\frac{1}{5}} \div \bigg(\dfrac{2}{3} \bigg)^{2}\Bigg]}} \\[/tex]

We can write as :

27 = 3 × 3 × 3 = 3³

8 = 2 × 2 × 2 = 2³

243 = 3 × 3 × 3 × 3 × 3 = 3⁵

32 = 2 × 2 × 2 ×2 × 2 = 2⁵

[tex]\sf{\longmapsto{\bigg( \dfrac{3 \times 3 \times 3}{2 \times 2 \times 2} \bigg)^{\frac{1}{3}} \times \Bigg[\bigg( \dfrac{3 \times 3 \times 3 \times 3 \times 3}{2 \times 2 \times 2 \times 2 \times 2} \bigg)^{\frac{1}{5}} \div \bigg(\dfrac{2}{3} \bigg)^{2}\Bigg]}} \\[/tex]

[tex]\sf{\longmapsto{\bigg( \dfrac{{(3)}^{3}}{{(2)}^{3}} \bigg)^{\frac{1}{3}} \times \Bigg[\bigg( \dfrac{({3}^{5})}{{(2)}^{5}} \bigg)^{\frac{1}{5}} \div \bigg(\dfrac{2}{3} \bigg)^{2}\Bigg]}} \\[/tex]

Now, we can write as :

(3³/2³) = (3/2)³

(3⁵/2⁵) = (3/2)⁵

[tex]\sf{\longmapsto{\left\{\bigg(\frac{3}{2} \bigg)^{3} \right\}^{\frac{1}{3}} \times \Bigg[\left\{\bigg(\frac{3}{2} \bigg)^{5} \right\}^{\frac{1}{5}} \div \bigg(\dfrac{2}{3} \bigg)^{2}\Bigg]}} \\[/tex]

Now using law of exponent :

[tex]{\sf{({a}^{m})^{n} = {a}^{mn}}}[/tex]

[tex]\sf{\longmapsto{\bigg( \frac{3}{2} \bigg)^{3 \times \frac{1}{3}} \times \Bigg[\bigg(\frac{3}{2} \bigg)^{5 \times \frac{1}{5}} \div \bigg(\dfrac{2}{3} \bigg)^{2}\Bigg]}} \\[/tex]

[tex] \sf{\longmapsto{\bigg( \frac{3}{2} \bigg)^{\frac{3}{3}} \times \Bigg[\bigg(\frac{3}{2} \bigg)^{\frac{5}{5}} \div \bigg(\dfrac{2}{3} \bigg)^{2}\Bigg]}} \\[/tex]

[tex]\sf{\longmapsto{\bigg( \frac{3}{2} \bigg)^{1} \times\Bigg[\bigg(\frac{3}{2} \bigg)^{1} \div \bigg(\dfrac{2}{3} \bigg)^{2}\Bigg]}} \\[/tex]

[tex]\sf{\longmapsto{\bigg( \frac{3}{2} \bigg)^{1} \times \Bigg[\bigg(\frac{3}{2} \bigg)^{1} \times \bigg(\dfrac{3}{2} \bigg)^{2}\Bigg]}} \\[/tex]

[tex]\sf{\longmapsto{\bigg( \frac{3}{2} \bigg)^{1} \times \Bigg[\bigg(\frac{3}{2} \bigg)^{1} \times \bigg(\dfrac{3}{2} \times \dfrac{3}{2} \bigg)\Bigg]}} \\[/tex]

[tex]\sf{\longmapsto{\bigg( \dfrac{3}{2} \bigg)^{1} \times \Bigg[\bigg(\dfrac{3}{2} \bigg)^{1} \times \bigg(\dfrac{3 \times 3}{2 \times 2}\bigg)\Bigg]}} \\[/tex]

[tex]\sf{\longmapsto{\bigg( \dfrac{3}{2} \bigg)^{1} \times \Bigg[\bigg(\dfrac{3}{2} \bigg)^{1} \times \bigg(\dfrac{9}{4}\bigg)\Bigg]}} \\[/tex]

[tex] \sf{\longmapsto{\bigg( \frac{3}{2} \bigg)\times \Bigg[\bigg(\frac{3}{2} \bigg)\times \bigg(\dfrac{9}{4}\bigg)\Bigg]}} \\[/tex]

[tex]\sf{\longmapsto{\bigg( \dfrac{3}{2} \bigg)\times \Bigg[ \: \: \dfrac{3}{2} \times \dfrac{9}{4} \: \: \Bigg]}}\\[/tex]

[tex]\sf{\longmapsto{\bigg( \dfrac{3}{2} \bigg)\times \Bigg[ \: \: \dfrac{3 \times 9}{2 \times 4} \: \: \Bigg]}} \\[/tex]

[tex]\sf{\longmapsto{\bigg(\dfrac{3}{2} \bigg)\times \Bigg[ \: \: \dfrac{27}{8} \: \: \Bigg]}} \\[/tex]

[tex]\sf{\longmapsto{\dfrac{3}{2} \times \dfrac{27}{8}}} \\[/tex]

[tex]\sf{\longmapsto{\dfrac{3 \times 27}{2 \times 8}}} \\[/tex]

[tex] \sf{\longmapsto{\dfrac{81}{16}}\: ≈ \:5.0625\:\red{Ans.}} \\[/tex]

9514 1404 393

Answer:

  7

Step-by-step explanation:

Of the 9 2-digit numbers ending in 6, only 2 are perfect squares: 16 and 36. The other 7 are not perfect squares.

Answer:

-5/6a - 5/8

Step-by-step explanation:

i took the k12 quiz

a is right its 58

and in my opinion i think b would be 58*5=290 cents, since nickel is worth 5 cents.

but gotta check other people's answers

Answer:

C

Step-by-step explanation:

Answer:

A) No solution

Step-by-step explanation:

Given the systems of linear equations, y = 2x + 1 and y = 2x - 1:

Both equations in the system have the same slope, m = 2, thus forming parallel lines.  Since their lines are parallel from each other, then it means that their lines will never intersect.

Therefore, the given systems of linear equation is an inconsistent system that has no solution.

JK=JH

[tex]\\ \sf\longmapsto 10t=7t+15[/tex]

[tex]\\ \sf\longmapsto 10t-7t=15[/tex]

[tex]\\ \sf\longmapsto 3t=15[/tex]

[tex]\\ \sf\longmapsto t=15/3=5[/tex]

Answer:

[tex]y=-\frac{1}{4}x\\[/tex]

Step-by-step explanation:

The slope is calculated by "up ÷ across".

= -1 ÷ 4

= [tex]-\frac{1}{4}[/tex]

The y-intercept is just 0 (because the line meets at the y axis at 0).

So, using [tex]y=mx+b[/tex] (where m = slope and b = y-intercept),

[tex]y=-\frac{1}{4}x+0[/tex]

but the '+0' is unnecessary so we just say [tex]y=-\frac{1}{4}x[/tex]

Answer:

£86.50

Step-by-step explanation:

To find how much 10 t-shirts cost we have to find how much they cost individually.

3 T- Shirts / 25.95 = 8.65 which means that each T-Shirt Costs 8.65.

8.65 x 10 T-Shirts = £86.50

Answer:

uh.. shii i dont know

Step-by-step explanation:

∆HFE is similar to ∆HFG and ∆ HEG (they all are right triangles and have the same, shape.)

The given triangle ΔHFE is similar to triangle GFH and triangle GHE.

Similar triangles are two triangles that have the same shape but may differ in size. They have corresponding angles that are congruent and corresponding sides that are proportional.

Here,

Since triangle HFE is similar to triangle GFH,
∠HEF = ∠GHF,
∠FHE = ∠HGF, and ∠HFE = ∠HFG

Similarly

ΔHFE is similar to triangle ΔGHE.

Similar triangles are useful in many areas of mathematics and in the real world, such as in architecture, engineering, and cartography, where they are used to make scale models of buildings and other structures.

Thus, the given triangle ΔHFE is similar to triangle GFH and triangle GHE.

Learn more about similar triangles here:

https://brainly.com/question/12101336
#SPJ2

Answer:

I can't read that...........