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3 years ago

Helpppppp meeeeeee plz

Mathematics

2 answers:

V125BC [204]3 years ago

6 0

60 total fourth graders and 43 are girls. To find the number of boys, subtract.

60 - 43 = 17 boys

If 5 girls were absent, subtract.

43 - 5 = 38 girls present

If 4 boys were absent, subtract.

17 - 4 = 13 boys present

Therefore, there were 13 boys present.

Best of Luck!

photoshop1234 [79]3 years ago

4 0

Answer:

13 boys were present

Step-by-step explanation:

→ In total there is 60 fourth graders at the Small Tree School.

→ 43 of them are girls and this information is given to us which means that there is 17 boys because 60 - 43 = 17

→ We want to know know how many boys were at the school on Friday considering that 5 fourth grade girls and 4 fourth grade boys were absent

→ We know that there is a total of 17 boys and 4 fourth grade boys were absent so 17 - 4 will give us the number of boys that were present on Friday so 17 - 4 = 13. 13 boys were present considering 5 fourth grade girls and 4 fourth grade boys were absent

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Simplify the expression (4 + 5i)(4 - 5i). 16 - 20i 16 + 20i -9 41

AnnZ [28]

Answer:

The answer i got was 41. i just took a test with the same problem and it was correct. Hope this helps.

Step-by-step explanation:

3 0

3 years ago

Solve the following using Substitution method 2x – 5y = -13 3x + 4y = 15

Digiron [165]

$\huge \boxed{\mathfrak{Question} \downarrow}$

Solve the following using Substitution method

2x – 5y = -13

3x + 4y = 15

$\large \boxed{\mathfrak{Answer \: with \: Explanation} \downarrow}$

$\left. \begin{array} { l } { 2 x - 5 y = - 13 } \\ { 3 x + 4 y = 15 } \end{array} \right.$

To solve a pair of equations using substitution, first solve one of the equations for one of the variables. Then substitute the result for that variable in the other equation.

$2x-5y=-13, \: 3x+4y=15$

Choose one of the equations and solve it for x by isolating x on the left-hand side of the equal sign. I'm choosing the 1st equation for now.

$2x-5y=-13$

Add 5y to both sides of the equation.

$2x=5y-13$

Divide both sides by 2.

$x=\frac{1}{2}\left(5y-13\right) \\$

Multiply $\frac{1}{2}\\$ times 5y - 13.

$x=\frac{5}{2}y-\frac{13}{2} \\$

Substitute $\frac{5y-13}{2}\\$ for x in the other equation, 3x + 4y = 15.

$3\left(\frac{5}{2}y-\frac{13}{2}\right)+4y=15 \\$

Multiply 3 times $\frac{5y-13}{2}\\$ .

$\frac{15}{2}y-\frac{39}{2}+4y=15 \\$

Add $\frac{15y}{2} \\$ to 4y.

$\frac{23}{2}y-\frac{39}{2}=15 \\$

Add $\frac{39}{2}\\$ to both sides of the equation.

$\frac{23}{2}y=\frac{69}{2} \\$

Divide both sides of the equation by 23/2, which is the same as multiplying both sides by the reciprocal of the fraction.

$\large \underline{ \underline{ \sf \: y=3 }}$

Substitute 3 for y in $x=\frac{5}{2}y-\frac{13}{2}\\$ . Because the resulting equation contains only one variable, you can solve for x directly.

$x=\frac{5}{2}\times 3-\frac{13}{2} \\$

Multiply 5/2 times 3.

$x=\frac{15-13}{2} \\$

Add $-\frac{13}{2}\\$ to $\frac{15}{2}\\$ by finding a common denominator and adding the numerators. Then reduce the fraction to its lowest terms if possible.