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Dvinal [7]
3 years ago
13

Mandy spent $75 and earned $75 number line model

Mathematics
1 answer:
damaskus [11]3 years ago
6 0
X-75+75= x

Is a model that could be used to anaslay the question
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Lines AC←→ and DB←→ intersect at point W. Also, m∠DWC=138° .
mr_godi [17]

m∠DWC=138°, ∠AWB = 138°, ∠AWD = 42°, ∠BWC = 42°

Solution:

Line \overrightarrow{A C} \text { and } \overrightarrow{B D} intersect at a point W.

Given m \angle D W C=138^{\circ}.

<em>Vertical angle theorem:</em>

<em>If two lines intersect at a point then vertically opposite angles are congruent.</em>

<u>To find the measure of all the angles:</u>

∠AWB and ∠DWC are vertically opposite angles.

Therefore, ∠AWB = ∠DWC

⇒ ∠AWB = 138°

Sum of all the angles in a straight line = 180°

⇒ ∠AWD + ∠DWC = 180°

⇒ ∠AWD + 138° = 180°

⇒ ∠AWD = 180° – 138°

⇒ ∠AWD = 42°

Since ∠AWD and ∠BWC are vertically opposite angles.

Therefore, ∠AWD = ∠BWC

⇒ ∠BWC = 42°

Hence the measure of the angles are

m∠DWC=138°, ∠AWB = 138°, ∠AWD = 42°, ∠BWC = 42°.

7 0
3 years ago
For a car moving at a constant speed, the distance traveled varies directly with the time spent driving. If such a car travels 2
ipn [44]

Answer:

120 miles.

Step-by-step explanation:

Use long division to find how many miles the car travels in 1 hour then we multiply by 4

: (7÷210=30)×4=120miles. Hope this helped ;D

6 0
3 years ago
A rectangular parking lot has an area of 15,000 feet squared, the length is 20 feet more than the width. Find the dimensions
faust18 [17]

Dimension of rectangular parking lot is width = 112.882 feet and length = 132.882 feet

<h3><u>Solution:</u></h3>

Given that  

Area of rectangular parking lot = 15000 square feet

Length is 20 feet more than the width.

Need to find the dimensions of rectangular parking lot.

Let assume width of the rectangular parking lot in feet be represented by variable "x"

As Length is 20 feet more than the width,

so length of rectangular parking plot = 20 + width of the rectangular parking plot

=> length of rectangular parking plot = 20 + x = x + 20

<em><u>The area of rectangle is given as:</u></em>

\text {Area of rectangle }=length \times width

Area of rectangular parking lot = length of rectangular parking plot \times width of the rectangular parking

\begin{array}{l}{=(x+20) \times (x)} \\\\ {\Rightarrow \text { Area of rectangular parking lot }=x^{2}+20 x}\end{array}

But it is given that Area of rectangular parking lot = 15000 square feet

\begin{array}{l}{=>x^{2}+20 x=15000} \\\\ {=>x^{2}+20 x-15000=0}\end{array}

Solving the above quadratic equation using quadratic formula

<em><u>General form of quadratic equation is  </u></em>

{ax^{2}+\mathrm{b} x+\mathrm{c}=0

And quadratic formula for getting roots of quadratic equation is

x=\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}

In our case b = 20, a = 1 and c = -15000

Calculating roots of the equation we get

\begin{array}{l}{x=\frac{-(20) \pm \sqrt{(20)^{2}-4(1)(-15000)}}{2 \times 1}} \\\\ {x=\frac{-(20) \pm \sqrt{400+60000}}{2 \times 1}} \\\\ {x=\frac{-(20) \pm \sqrt{60400}}{2}} \\\\ {x=\frac{-(20) \pm 245.764}{2 \times 1}}\end{array}

\begin{array}{l}{=>x=\frac{-(20)+245.764}{2 \times 1} \text { or } x=\frac{-(20)-245.764}{2 \times 1}} \\\\ {=>x=\frac{225.764}{2} \text { or } x=\frac{-265.764}{2}} \\\\ {=>x=112.882 \text { or } x=-132.882}\end{array}

As variable x represents width of the rectangular parking lot, it cannot be negative.

=> Width of the rectangular parking lot "x" = 112.882 feet  

=> Length of the rectangular parking lot = x + 20 = 112.882 + 20 = 132.882

Hence can conclude that dimension of rectangular parking lot is width = 112.882 feet and length = 132.882 feet.

3 0
3 years ago
Karina needs a total of $45 to buy her mother a birthday present. She has saved 20% of the amount so far. How much has she saved
professor190 [17]

Answer:

$9.00

Step-by-step explanation:

$45.00x20%=9

7 0
3 years ago
Read 2 more answers
Find the gradient of this line. Use gradient = rise/run
Marysya12 [62]

Step-by-step explanation:

the gradient of the line = 2 units/ 3 units = 2/3

5 0
3 years ago
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