Simplify 3a+4+b+2a+5+b

Mathematics

2 answers:

Alinara [238K]3 years ago

8 0

Answer: 5 a + 2 b + 9

Step-by-step explanation: Hope this helped!!!!

butalik [34]3 years ago

3 0

Answer:

Step-by-step explanation: you add 3a and 2a because they have the same variable, then you add 4 and 5, then you add b and b, to solve this you have to find like terms.

Answer: 5a+2b+9

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katovenus [111]

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7 0

3 years ago

Ginny's piggybank contained $8.80 in quarters, dimes, and nickels. there were two more than five times as many nickels as quarte

SSSSS [86.1K]

Answer:

There are 13 quarters, 67 nickels, and 22 dimes.

Step-by-step explanation:

The value of a quarter is $0.25, the value of dime is $0.10, and the value of a nickel is $0.05. Let there be q quarters, n nickels, and d dimes.

Let's write equations from the statements given:

"Ginny's piggybank contained $8.80 in quarters, dimes, and nickels.":

$0.25q+0.1d+0.05n=8.80$

"there were two more than five times as many nickels as quarters":

$5q+2=n$

"4 less than twice as many dimes as quarters":

$2q-4=d$

Plugging in equation 2 and 3 into equation 1, we can solve for q:

$0.25q+0.1(2q-4)+0.05(5q+2)=8.80\\0.25q+0.2q-0.4+0.25q+0.1=8.80\\0.7q=8.80+0.4-0.1\\0.7q=9.1\\q=\frac{9.1}{0.7}=13$

Substituting q = 13 into equation 2 will give us nickels:

$n=5q+2\\n=5(13)+2\\n=67$

Substituting q = 13 into equation 3 will give us dimes:

$d=2q-4\\d=2(13)-4\\d=22$

There are 13 quarters, 67 nickels, and 22 dimes.

4 0

3 years ago

HELP PLEASE!!!!! Solve each system.

Ede4ka [16]

$15.\ \text{Elimination method}\\\underline{+\left\{\begin{array}{ccc}y=-3x+9\\y=3x+3\end{array}\right}\qquad\text{add both sides of the equations}\\.\qquad2y=12\qquad\text{divide both sides by 2}\\.\qquad \boxed{y=6}\\\\\text{put the value of y to the second equation}\\\\6=3x+3\qquad\text{subtract 3 from both sides}\\3=3x\qquad\text{divide both sides by 3}\\1=x\to\boxed{x=1}\\\\Answer:\ x=1,\ y=6\to(1,\ 6)$

$16.\ \text{Substitution method}\\\left\{\begin{array}{ccc}y=11x+4\\y=7x-16\end{array}\right\Rightarrow11x+4=7x-16\\\\11x+4=7x-16\qquad\text{subtract 4 from both sides}\\11x=7x-20\qquad\text{subtract 7x from both sides}\\4x=-20\qquad\text{divide both sides by 4}\\\boxed{x=-5}\\\\\text{Put the value of x to the first equation}\\\\y=11(-5)+4\\y=-55+4\\\boxed{y=-51}\\\\Answer:\ x=-5,\ y=-51\to(-5,\ -51)$