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4 years ago

8

Z roms

1 answer:

andre [41]4 years ago

8 0

Answer:

answer is $28.87

Step-by-step explanation:

You might be interested in

The dimensions of a rectangle are such that its length is 3 in. more than its width. If the length were doubled and if the width

Kay [80]

Answer:

The length of the rectangle is 11 inch and its width is 8 inch.

Step-by-step explanation:

Given:

Rectangle $(R_1)$ and rectangle $(R_2)$ .

Let the width of the rectangle be 'x' inch.

Its length = '(x+3)' inch

Area of the rectangle = $(A_1)$

⇒ $A_1$ = $(x)(x+3)$

Now for rectangle $(R_2)$ its width is decreased by 1 inch and length is doubled,the area $(A_2)$ will increased by 66 sq-inch.

So,

$A_2=A_1+66$

According to the question.

⇒ Area of rectangle $(R_2)$ with width '(x-1)' and length '2(x+3)'='(2x+6)' .

⇒ $A_2=(x-1)(2x+6)$

Plugging the values of $A_2$ in terms of $A_1$ .

⇒ $A_1+66= (x-1)(2x+6)$

Plugging $A_1$ = $(x)(x+3)$ into the equation.

⇒ $(x)(x+3)+66=(x-1)(2x+6)$

⇒ $x^2+3x+66=2x^2+6x-2x-6$

⇒ $x^2-2x^2+3x-6x+2x+6+66=0$

⇒ $-x^2-x+72=0$

Now using quadratic formula we can find the value of 'x'.

Quadratic formula, (x)= $\frac{\pm b+\sqrt{b^2-4ac} }{2a}$

x= -9 and x= 8

Discarding the negative value.

Our width = 'x' = 8 inches.

So the length of the rectangle = 11 inches and width of the rectangle = 8 inches.

7 0

3 years ago

Last year, the City of Greenpoint experienced a budget deficit of $ 838,880. By comparison, the expected deficit for this year i

Arturiano [62]

Here I will help you

Step-by-step explanation:

We have to divide 838,880 in to the 50% Like this

838,880/50%=1676

4 0

3 years ago

Polygon JKLM is dilated by a scale factor of 2.5 with point C as the center of dilation, resulting in the image J′K′L′M′. If poi

user100 [1]

I think the slope of L'M' is still 1.75 and it still passes through point C.

5 0

3 years ago

Idek what this is lol

STatiana [176]

Answer:

lol means laugh out loud

5 0

4 years ago

Which of the following is the graphed below

choli [55]

Answer:

$A.\:y=\left \{ {{x^3-3,x\le2} \atop {x^2+6,\:x>2}} \right.$

Step-by-step explanation:

The quadratic function has equation $y=x^2+6$ and the cubic function has equation $x^3-3$ .

The function $y=x^2+6$ is defined on the interval $x\:>\:2$ .

The function $y=x^3-3$ is defined on the interval $x\:\le 2$

The required piece-wise function is

$y=\left \{ {{x^3-3,x\le2} \atop {x^2+6,\:x>2}} \right.$

8 0

4 years ago