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Lana71 [14]
3 years ago
10

The gym has a total of 25 treadmills and stationary bikes. there are seven more stationary bikes than treadmills.

Mathematics
1 answer:
Rudik [331]3 years ago
3 0

The system of equations is t+b=25\\b=t+7

Step-by-step explanation:

We can answer this question as follows.

First of all, we call:

t = number of treadmills

b = number of stationary bikes

The two conditions that we have can be translated into equations as follows:

- The gym has a total of 25 treadmills and stationary bikes:

t+b=25

- There are seven more stationary bikes than treadmills:

b=t+7

So the system of equations to solve is

t+b=25\\b=t+7

We now solve it in the following way: first, we rewrite the second equation by bringing t on the left side,

t+b=25\\b-t=7

Now we add the 1st equation to the 2nd equation:

(t+b)+(b-t)=25+7\\t+b+b-t=32\\2b=32\\b=\frac{32}{2}=16

And therefore,

t+b=25\\t+16=25\\t=25-16=9

So, there are 16 stationary bikes and 9 treadmills.

Learn more about systems of equations:

brainly.com/question/13168205

brainly.com/question/3739260

#LearnwithBrainly

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Step-by-step explanation:

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y = 18 works because it is less than 22. (:

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Find the lengths and slopes of the diagonals to name the parallelogram. Choose the most specific name. E (-2, -4), F(0, -1), G(-
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Answer:

1) d) Square

2) Proofs that PWRS is a rhombus are

Length of QS ≠ PR and

Slope of segment QR and PS is -1/2 and Slope of segment RS and QP is -2.

Step-by-step explanation:

The given points (x, y) of the parallelogram are;

E(-2, -4), F(0, -1), G(-3, 1), H(-5, -2)

The slope, m, of the segments are found as follows;

Slope, \, m =\dfrac{y_{2}-y_{1}}{x_{2}-x_{1}}

By computation, the slope of segment EF = 1.5

The slope of segment FG = -0.67

The slope of segment GH = 1.5

The slope of segment HE = -0.67

Therefore, EF is parallel to GH and FG is parallel to HE

The length of the sides are;

\sqrt{\left (y_{2}-y_{1}  \right )^{2}+\left (x_{2}-x_{1}  \right )^{2}}

By computation, the length of segment EF = 3.61

The length of segment FG = 3.61

The length of segment GH = 3.61

The length of segment HE = 3.61

The diagonals are;

EG and FH

The length of segment EG = 5.099

The length of segment FH = 5.099

Therefore, the diagonals are equal and the parallelogram is a square

2) The given dimensions are;

P(-1, 3), Q(-2, 5), R(0, 4), S(1, 2)

A rhombus has all sides equal

The length of segment PQ = 2.24

The length of segment QR = 2.24

The length of segment RS = 2.24

The length of segment PS = 2.24

The diagonals are;

QS and PR

The length of segment QS = 4.24

The length of segment PR = 1.41

The slope of segment QR = -0.5

The slope of segment PS = -0.5

The slope of segment RS = -2

The slope of segment QP = -2

Therefore, QS≠QR the parallelogram is a rhombus

The correct option ;

Length of QS ≠ PR and

Slope of segment QR and PS is -1/2 and Slope of segment RS and QP is -2.

Where there are acute angles in parallelogram PQRS, then the correct option is d) Length of QR and PS is 2.2 and Length of RS and QP is 2.2

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