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Marysya12 [62]
3 years ago
15

Erica has 7 coins worth $1.55. If the coins are all quarters and nickels, how many quarters and nickels does she have?

Mathematics
1 answer:
Levart [38]3 years ago
5 0

Answer:

She has 6 quarters and 1 nickel

Step-by-step explanation:

6 quarters = $1.50

1 nickel = 5c

add those together you get $1.55

and only 7 coins

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(3x- 1) 71 solve for the equation and x please
Ksivusya [100]

Answer:

x=24

Step-by-step explanation:

The opposite angles are equal as per certain theorems.

3x-1=71

3x=72

x=24

4 0
3 years ago
The ratio of students to teachers is 2:3 respectively.If there are 21 teachers, how many students are there?
yulyashka [42]

Answer:14


Step-by-step explanation:

2:3

14:21

in 14 their is 2 7's

in 21 their is 4 7's


6 0
3 years ago
Which is true about the polynomial –8m3 + 11m?
Dima020 [189]

Answer:

Binomial with a degree of 3

Step-by-step explanation:

-8m^3 + 11m....notice that it has 2 terms....(-8m^3) and (11m). Having 2 terms makes it a binomial...if it would have had 3 terms, it would have been a trinomial. If it has only one variable, the degree is the highest exponent...so this has a degree of 3 since ^3 is the highest exponent.

so the answer is : binomial with a degree of 3

7 0
3 years ago
Read 2 more answers
Is x = 7 a solution to the equation 3x - 4 = 15? No Yes
Elis [28]

Answer:

No, because 3 x 7 = 21 - 4 = 17

Step-by-step explanation:

8 0
3 years ago
Read 2 more answers
1)A System of equations is shown below . What is the solution to the system of equations? 5x+2y=-15 2x-2y=-6
meriva

Answer:

x= -3 and y= 0

Step-by-step explanation:

5x+2y=-15

<u>2x-2y=-6     </u>

<u>7x        =-21</u>

x= -3

Putting value of x in equation 1  

5(-3) +2y=-15

-15+2y= -15

2y= 0

y= 0

This can be solved with the help of matrices

In matrix form the above equations can be written in the form

\left[\begin{array}{ccc}5&2\\2&-2\/\end{array}\right]  \left[\begin{array}{ccc}x\\y\\\end{array}\right]  = \left[\begin{array}{ccc}-15\\-6\\\end{array}\right]

Let

\left[\begin{array}{ccc}5&2\\2&-2\/\end{array}\right] = A  \left[\begin{array}{ccc}x\\y\\\end{array}\right]  = X  and  \left[\begin{array}{ccc}-15\\-6\\\end{array}\right]= B

Then AX= B

or X= A⁻¹ B

where  A⁻¹= adj A/ ║A║   where mod A≠ 0

adj A=  \left[\begin{array}{ccc}-2&-2\\-2&5\/\end{array}\right]

║A║= ( 5*-2- 2*2)= -10-4= -14≠0

X= A⁻¹ B

 \left[\begin{array}{ccc}x\\y\\\end{array}\right]    =- 1/14  \left[\begin{array}{ccc}-2&-2\\-2&5\/\end{array}\right]   \left[\begin{array}{ccc}-15\\-6\\\end{array}\right]

 \left[\begin{array}{ccc}x\\y\\\end{array}\right]    =- 1/14     \left[\begin{array}{ccc}-2*-15&+ -2*-6\\-2*-15&+ 5*-6\\\end{array}\right]

 \left[\begin{array}{ccc}x\\y\\\end{array}\right]  =- 1/14 \left[\begin{array}{ccc} 30&+12\\30&+-30\\\end{array}\right]

 \left[\begin{array}{ccc}x\\y\\\end{array}\right]  =- 1/14 \left[\begin{array}{ccc}42\\0\\\end{array}\right]

\left[\begin{array}{ccc}x\\y\\\end{array}\right]  = \left[\begin{array}{ccc}-42/14\\0/-14\\\end{array}\right]

\left[\begin{array}{ccc}x\\y\\\end{array}\right]  = \left[\begin{array}{ccc}-3\\0\\\end{array}\right]

From here x= -3 and y= 0

Solution Set = [(-3,0)]

3 0
3 years ago
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