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3 years ago

What is the product of seven and eight decreased by “x” if x =15?

Mathematics

2 answers:

jekas [21]3 years ago

7 0

Answer:

Step-by-step explanation:

product is another word for +

decreased is another word for -

____________________________

(7 + 8) - x

(7+8) - 15

15 - 15

= 0

yKpoI14uk [10]3 years ago

3 0

Answer:

Step-by-step explanation:

<em>Vocab:</em>

Product: Multiplication

Decrease: Subtraction

(7*8)-x

Substitute:

(7*8)-15

Multiply:

(56)-15

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Tom finished 1/3 of the 30 problems assigned in class. Steven did 3/10 of the problems. Juan finished 20% of the problems. Marcu

RoseWind [281]

Answer: Tom

Step-by-step explanation:

Tom did 1/3 of 30= 10 out of 30

Steven did 3/10 of 30= 9 out of 30

Juan did 20/100 of 30= 6 out of 30

Marcus did 8 out of 30

6 0

3 years ago

Read 2 more answers

The table shows the number of students in a class who play on a school team and/or are members of a school club:

eimsori [14]

This one is very simple.

create a simple equation to find out

15 = 12 + x

subtract 12 from each side

x= 3

the missing number is 3

or, to be even simpler, take the total, subtract the total from the number you know and the difference is the missing number.

hope I helped

ALSO, we can check going down vertically. 3 + 10 =13. this is the total listed at the bottom.

3 0

3 years ago

Approximate the integral integral integral integral f(x, y) dA by dividing the rectangle R with vertices (0, 0), (4, 0), (4, 2),

amm1812

Answer:

Step-by-step explanation:

Approximate the integral $\int\int\limits_R {f(x,y)} \, dA$ by dividing the region $R$ with vertices (0,0),(4,0),(4,2) and (0,2) into eight equal squares.

Find the sum $\sum\limits^8_{i=1}f(x_i,y_i)\delta A_i$

Since all are equal squares, so $\delta A_i=1$ for every $i$

$\sum\limits^8_{i=1}f(x_i,y_i)\delta A_i=f(x_1,y_1)\delta A_1+f(x_2,y_2)\delta A_2+f(x_3,y_3)\delta A_3+f(x_4,y_4)\delta A_4+f(x_5,y_5)\delta A_5+f(x_6,y_6)\delta A_6+f(x_7,y_7)\delta A_7+f(x_8,y_8)\delta A_8\\\\=f(0.5,0.5)(1)+f(1.5,0.5)(1)+f(2.5,0.5)(1)+f(3.5,0.5)(1)+f(0.5,1.5)(1)+f(1.5,1.5)(1)+f(2.5,1.5)(1)+f(3.5,1.5)(1)\\\\=0.5+0.5+1.5+0.5+2.5+0.5+3.5+0.5+0.5+1.5+1.5+1.5+2.5+1.5+3.5+1.5\\\\=24$

Thus, $\sum\limits^8_{i=1}f(x_i,y_i)\delta A_i=24$

Evaluating the iterate integral $\int\limits^4_0 \int\limits^2_0 {(x+y)} \, dydx=\int\limits^4_0 {[xy+\frac{y^2}{2} ]}\limits^2_0 \, dx =\int\limits^4_0 {[2x+2]}dx\\\\=[x^2+2x]\limits^4_0=24.$