All categories

4 years ago

Help me plzzzz!!!!! I need help 5-11

Mathematics

1 answer:

Gnom [1K]4 years ago

5 0

Well 5 - 11 is -6 so there is your anser

You might be interested in

Put the equation below in the form ax2 + bx + c = 0. Enter exponents using the caret ( ^ ). For example, you would enter 4x2 as

Ymorist [56]

X2+ 4x + 7 = 0 is the correct answer

4 0

3 years ago

Rhonda has $4.85 in coins. If she has six more nickels than dimes and twice as many quarters as dimes, how many coins of each ty

Novay_Z [31]

Answer: nickels = 13, dimes = 7, quarters = 14

<u>Step-by-step explanation:</u>

First, set up the equations for each coin:

$\begin {array}{c|c|c|l}\underline{\quad Type\quad }&\underline{Value}&\underline{Quantity}&\underline{\qquad \qquad \qquad Equation\qquad }\\ Nickel& \$0.05&6+d&0.05(6+d) = 0.30 + 0.05d\\Dime& \$0.10&d&\qquad 0.10(d)=0.10d\\Quarter& \$0.25&2d&\qquad 0.25(2d)=0.50d\\\end{array}$

Next, the sum of the coins is $4.85 so substitute and solve for the variable:

Nickels + Dimes + Quarters = Sum

(0.30 + 0.05d) + 0.10d + 0.50d = 4.85

0.30 + 0.65d = 4.85

0.65d = 4.55

d = 7

Lastly, plug the d-value into the quantity equation for the nickels and quarters to find their quantity.

nickels: 6 + d = 6 + (7) = 13

quarters: 2d = 2(7) = 14

6 0

3 years ago

PLEASE HELP!!!!!!!!!!!!!!!

gavmur [86]

Answer:

Step-by-step explanation:

because it the 0 that need to subtract

7 0

3 years ago

Jane buys 220 milliliters of lotion for $7.50. She divides the lotion evenlyamount 5 smaller bottles. How many liters are in eac

KonstantinChe [14]

220 (divide sign) 5 = 44ml (change it to liters) 0.044 (move the decimal from 44.0 their by two spots so it would be 0.044)

5 0

3 years ago

Read 2 more answers

HEEELLLPPP!!!!!!!!! PLEEAASEEE!!!!!!!!

Ksivusya [100]

Answer:

Step-by-step explanation:

$17. (a^{m})^{n}=a^{m*n}\\\\a^{m}*a^{n}=a^{m+n}\\\\-3k^{2}*(4k^{5})^{3}=-3k^{2}*4^{3}*k^{5*3}\\\\ = -3k^{2}*64*k^{15}\\\\= (-3)*64*k^{2+15}\\\\= - 192k^{17}$

$19. \dfrac{a^{m}}{a^{n}}=a^{m-n}\\\\\\\dfrac{(-7p^{4})*(-8p^{5})}{2p^{3}}= \dfrac{(-7)*(-8)*p^{4}*p^{5}}{2p^{3}}\\\\\\=(-7)*(-4)*p^{4+5-3} \\\\= 28p^{6}\\$

$20)\dfrac{15a^{16}b^{11}}{(3a^{4}b^{2})^{3}}=\dfrac{15a^{16}b^{11}}{3^{3}*a^{4*3}*b^{2*3}}\\\\\\=\dfrac{15a^{16}b^{11}}{27*a^{12}*b^{6}}\\\\=\dfrac{5*a^{16-12}*b^{11-6}}{9}\\\\\\=\dfrac{5a^{4}b^{5}}{9}$

4 0

2 years ago