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jarptica [38.1K]

2 years ago

15

Two workers produced 86 parts and the first one produced 8 more parts than the second one. How many parts did each worker produc

e?

2 answers:

Diano4ka-milaya [45]2 years ago

6 0

1st worker produced x pants,
2nd worker produced (x+8) pants.

x+x+8=86
2x=78
x= 39

x+8 = 39+8 = 47

1st worker produced 39 pants,
2nd worker produced 47 pants.

In-s [12.5K]2 years ago

4 0

Answer:

u mean parts not pants

Step-by-step explanation:

and the 1st one made 47 parts and the second one made 39

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Unit 7 polygons & quadrilaterals homework 3: rectangles Gina Wilson answer key

Irina-Kira [14]

Answer:

In the image attached you can find the Unit 7 homework.

We need to findt he missing measures of each figure.

<h3>1.</h3>

Notice that the first figure is a rectangle, which means opposite sides are congruent so,

VY = 19

WX = 19

YX = 31

VW = 31

To find the diagonals we need to use Pythagorean's Theorem, where the diagonals are hypothenuses.

$VX^{2}=19^{2}+31^{2}\\ VX=\sqrt{361+961}=\sqrt{1322} \\VX \approx 36.36$

Also, $YW \approx 36.36$ , beacuse rectangles have congruent diagonals, which intercect equally.

That means, $ZX = \frac{VX}{2} \approx \frac{36.36}{2}\approx 18.18$

<h3>2.</h3>

Figure number two is also a rectangle.

If GH = 14, that means diagonal GE = 28, because diagonals intersect in equal parts.

Now, GF = 11, because rectangles have opposite sides congruent.

DF = 28, because in a reactangle, diagonals are congruent.

HF = 14, because its half of a diagonal.

To find side DG, we need to use Pythagorean's Theorem, where GE is hypothenuse

$GE ^{2}=11^{2}+DG^{2}\\28^{2}-11^{2}=DG^{2}\\DG=\sqrt{784-121}=\sqrt{663}\\ DG \approx 25.75$

<h3>3.</h3>

This figure is also a rectangle, which means all four interior angles are right, that is, equal to 90°, which means angle 11 and the 59° angle are complementary, so

$\angle 11 +59\°=90\°\\\angle 11=90\°-59\°\\\angle 11=31\°$

Now, angles 11 and 4 are alternate interior angles which are congruent, because a rectangle has opposite congruent and parallel sides.

$\angle 4 = 31\°$

Which means $\angle 3 = 59\°$ , beacuse it's the complement for angle 4.

Now, $\angle 6 = 59\°$ , because it's a base angle of a isosceles triangle. Remember that in a rectangle, diagonals are congruent, and they intersect equally, which creates isosceles triangles.

$\angle 9=180-59-59=62$ , by interior angles theorem.

$\angle 8 =62$ , by vertical angles theorem.

$\angle 10 = 180- \angle 9=180-62=118\°$ , by supplementary angles.

$\angle 7 = 118\°$ , by vertical angles theorem.

$\angle 5=90-59=31$ , by complementary angles.

$\angle 2 = \angle 5 = 31\°$ , by alternate interior angles.

$\angle 1 = 59\°$ , by complementary angles.

<h3>4.</h3>

$m\angle BCD=90\°$ , because it's one of the four interior angles of a rectangle, which by deifnition are equal to 90°.

$m\angle ABD = 6\° = m\angle BDC$ , by alternate interior angles and by given. $m\angle CBE=90-6=84$ , by complementary angles.

$m\angle ADE=90-6=84$ , by complementary angles.

$m\angle AEB=180-6-6=168\°$ , by interior angles theorem.

$m\angle DEA=180-168=12$ , by supplementary angles.

<h3>5.</h3>

$m\angle JMK=180-126=54$ , by supplementary angles.

$m\angle JKH=\frac{180-54}{2}=\frac{126}{2}=63$ , by interior angles theorem, and by isosceles triangle theorem.

$m\angle HLK=90\°$ , by definition of rectangle.

$m\angle HJL=\frac{180-126}{2}=27$ , by interior angles theorem, and by isosceles triangle theorem.

$m\angle LHK=90-27=63$ , by complementary angles.

$m\angle = JLK= m\angle HJL=27$ , by alternate interior angles.

<h3>6.</h3>

The figure is a rectangle, which means its opposite sides are equal, so

$WZ=XY\\7x-6=3x+14\\7x-3x=14+6\\4x=20\\x=\frac{20}{4}\\ x=5$

Then, we replace this value in the expression of side WZ

$WZ=7x-6=7(5)-6=35-6=29$

Therefore, side WZ is 29 units long.

<h3>7.</h3>

We know that the diagonals of a rectangle are congruent, so

$SQ=PR\\11x-26=5x+28\\11x-5x=28+26\\6x=54\\x=\frac{54}{6}\\ x=9$

Then,

$PR=5x+28=5(9)+28=45+28=73$

Therefore, side PR is 73 units long.

3 0

2 years ago

(8.218+9.93)+(17.782+0.07)

Licemer1 [7]

First, using the order of operations(PEMDAS), you would solve what is inside the parenthesis.
<span>(8.218 + 9.93) + (17.782 + 0.07)
(</span>18.148) + (17.852)
18.148 + 17.852

Now, all you would have to do is add the two sums.
18.148 + 17.852 = <span>36
</span>
The answer would be 36.

I hope this helps!

5 0

2 years ago

What are the zeros of the quadratic function f(x) = 6x^2 + 12x – 7?

strojnjashka [21]

Answer:

$x1=\frac{-2+\sqrt{26/3}}{2}$

$x2=\frac{-2-\sqrt{26/3} }{2}$

Step-by-step explanation:

To find the zeros of the quadratic function f(x)=6x^2 + 12x – 7 we need to factorize the polynomial.

To do so, we need to use the quadratic formula, which states that the solution to any equation of the form ax^2 + bx + c = 0 is:

$x=\frac{-b±\sqrt{b^{2}-4ac}}{2a}$

So, the first thing we're going to do is divide the whole function by 6:

6x^2 + 12x – 7 = 0 -> x^2 + 2x - 7/6

This step is optional, but it makes things quite easier.

Then we using the quadratic formula, where:

a=1, b= 2, c = -7/6.

Then:

$x=\frac{-2±\sqrt{2^{2}-4(1)(-7/6)}}{2}$

$x=\frac{-2±\sqrt{4 +14/3}}{2}$

$x=\frac{-2±\sqrt{26/3}}{2}$

So the zeros are:

$x1=\frac{-2+\sqrt{26/3}}{2}$

$x2=\frac{-2-\sqrt{26/3}}{2}$

4 0

2 years ago

A stadium brings in $16.25 million per year. it pays football-related expenses of $13.5 million and stadium expenses of $2.7 mil

Ganezh [65]

We have that

Profit margin is calculated by finding the net profit as a percentage of the revenue

Profit margin = [Net profit / Revenue]

Net profit= [Revenue-Cost]

we know that

Revenue = $16.25 million

Cost = $13.5 million + $2.7 million

Net profit = [16.25 million - (13.5 million + 2.7 million)]

Net profit = $0.05 million

Profit margin = 0.05 / 16.25

Profit margin = 0.003077 or 0.3077%

3 0

2 years ago

Match the following guess solutions yp for the method of undetermined coefficients with the second-order nonhomogeneous linear e

qaws [65]

Answer:

Step-by-step explanation:

For each of the yp(x) we can deduce the characteristic polynomial of the differential equation

A.

$yp(x)=Ax^2+Bx+C\\m^2=0\\y''=0$

B.

$yp(x)=Ae^{2x}\\(m-2)m=0\\m^2-2m=0\\y''-2y=0$

C.

$yp(x)=Acos2x+Bsin2x\\m_1=2i\\m_2=-2i\\y''+4=0$

D.

$yp(x)=(Ax+B)cos2x+(Cx+D)sin2x\\yp(x)=xcos4x$

hope this helps!!

8 0

2 years ago