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Katarina [22]
3 years ago
8

A rectangular room’s length is 2 feet less than its’ width and the area of the room is 120 ft^2

Mathematics
2 answers:
lisabon 2012 [21]3 years ago
7 0

Step-by-step explanation:

120/W=1

120/W= I

I =W is equal to 10

julia-pushkina [17]3 years ago
3 0

Answer:

a. 120/w = l

120/l = w

b. l times w = 120

Step-by-step explanation:

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When a decimal is written in word form, what indicates that the equivalent form is a mixed number and not a fraction
nignag [31]
If there is a number before the decimal point. So one point five six (1.56) would be have a mixed number of one before the normal fractions ; 1 56/100
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Solve for xxxx plsss​
natali 33 [55]

Answer:

Solution,

Here,∠FGD=∠CGE=30°[Being vertically opposite angles]

Now  we have,

90°+∠CGE+x=180°

or,90°+30°+x=180°

0r,x=180°-30°-90°

→x=60°

Step-by-step explanation:

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I need help on this question I need the answer please
SCORPION-xisa [38]
Answer: -1
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2 years ago
Using a graphing utility, find the exact solutions of the system. Round to the nearest hundredth and choose a solution to the sy
Margaret [11]

Answer:

Part 1) The exact solutions are

(\frac{-1+\sqrt{21}} {2},4+\sqrt{21})   and  (\frac{-1-\sqrt{21}} {2},4-\sqrt{21})

Part 2) (1.79, 8.58)

Step-by-step explanation:

we have

y=x^{2} +3x ----> equation A

y=2x+5 ----> equation B

we know that

When solving the system of equations by graphing, the solution of the system is the intersection points both graphs

<em>Find the exact solutions of the system</em>

equate equation A and equation B

x^{2} +3x=2x+5\\x^{2} +3x-2x-5=0\\x^{2} +x-5=0

The formula to solve a quadratic equation of the form

ax^{2} +bx+c=0

is equal to

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

in this problem we have

x^{2} +x-5=0  

so

a=1\\b=1\\c=-5

substitute in the formula

x=\frac{-1\pm\sqrt{1^{2}-4(1)(-5)}} {2(1)}

x=\frac{-1\pm\sqrt{21}} {2}

so

The solutions are

x_1=\frac{-1+\sqrt{21}} {2}

x_2=\frac{-1-\sqrt{21}} {2}

<em>Find the values of y</em>

<em>First solution</em>

For x_1=\frac{-1+\sqrt{21}} {2}

y=2(\frac{-1+\sqrt{21}} {2})+5

y=-1+\sqrt{21}+5\\\\y=4+\sqrt{21}

The first solution is the point (\frac{-1+\sqrt{21}} {2},4+\sqrt{21})

<em>Second solution</em>

For x_2=\frac{-1-\sqrt{21}} {2}

y=2(\frac{-1-\sqrt{21}} {2})+5

y=-1-\sqrt{21}+5\\\\y=4-\sqrt{21}

The second solution is the point (\frac{-1-\sqrt{21}} {2},4-\sqrt{21})

Round to the nearest hundredth

<em>First solution </em>

(\frac{-1+\sqrt{21}} {2},4+\sqrt{21}) -----> (1.79,8.58)

(\frac{-1-\sqrt{21}} {2},4-\sqrt{21}) -----> (-2.79,-0.58)

see the attached figure to better understand the problem

6 0
3 years ago
Find the missing length
natka813 [3]

Answer:

x = 45

Step-by-step explanation:

Corresponding sides of similar triangles will be in same ratio.

\dfrac{x}{x+18}=\dfrac{20}{28}

Cross multiply

x*28 = 20*(x+18)   {Use distributive method}

28x = 20x + 360

28x - 20x = 360

8x = 360

 x = 360/8

x = 45

3 0
2 years ago
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