A construction manager weighs a bundle of 9 identical bricks and a 7 pound concrete block. The bundle weighs 30 pounds.

Mathematics

1 answer:

denis23 [38]2 years ago

6 0

Answer:

30 = 9x + 7

Step-by-step explanation:

The total weight of the bundle = 30 pounds

The weight of the bricks = x pounds

But he carried 9 bricks = 9x pounds

He also carried additional 7 pounds of concrete block.

An equation to show the total weight he carried is

30 pounds = (9x + 7) pounds

Since 9x represents the total weight of the bricks he carried

7 represents the weight of the concrete block

And 30 pounds represents the total weight of everything he carried.

You might be interested in

720÷90=72 tens ÷ 9 tens = what?

Elden [556K]

720 divided by 72 is 80
7 tens divided by 9 is 8

6 0

2 years ago

1. If you wanted to make a triangle with side lengths of 3 units, 3 units, and 4

ivanzaharov [21]

Answer:

Since 4 is the largest number its the hypotenuse. The two 3s are equal so they are interchangeable so it doesn't matter where you put them. If its a right triangle the hypotenuse is opposite the right angle.

Step-by-step explanation:

7 0

2 years ago

Line k is parallel to y= -2x - 1 and passes through the point (4,3). What is the y intercept of line k?

Schach [20]

Yk what all above you can’t go wrong with choosing all of the em tbh cuz u would get one of them righ

4 0

1 year ago

What is the result of substituting for y in the bottom equation y=x+3 y=x^2+2x-4

8_murik_8 [283]

The solutions are (2.1925, 5.1925) and (-3.1925, -0.1925)

Solution:

Given that,

$y = x + 3 ------- eqn 1\\\\y = x^2 + 2x - 4 ----- eqn 2$

We have to substitute eqn 1 in eqn 2

$x + 3 = x^2 + 2x - 4$

$\mathrm{Switch\:sides}\\\\x^2+2x-4=x+3\\\\\mathrm{Subtract\:}3\mathrm{\:from\:both\:sides}\\\\x^2+2x-4-3=x+3-3\\\\\mathrm{Simplify}\\\\x^2+2x-7=x\\\\\mathrm{Subtract\:}x\mathrm{\:from\:both\:sides}\\\\x^2+2x-7-x=x-x\\\\\mathrm{Simplify}\\\\x^2+x-7=0$

$\mathrm{Solve\:with\:the\:quadratic\:formula}\\\\\mathrm{For\:a\:quadratic\:equation\:of\:the\:form\:}ax^2+bx+c=0\mathrm{\:the\:solutions\:are\:}\\\\x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}\\\\\mathrm{For\:}\quad a=1,\:b=1,\:c=-7\\\\x =\frac{-1\pm \sqrt{1^2-4\cdot \:1\left(-7\right)}}{2\cdot \:1}$