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abruzzese [7]
3 years ago
13

the member of the booster club decided to sell flowers they planned to charge 7 for a flower but they reduced the price by 2 the

sold 140 at the reduced price

Mathematics
2 answers:
11Alexandr11 [23.1K]3 years ago
7 0
B because it needs to be 140 multiplied by 5 or 7-2.
Grace [21]3 years ago
6 0

Answer:

B

Step-by-step explanation:

140*(7-2) shows $7 - $2 is $5

then, 140 flowers*$5 = $700

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Ivanshal [37]
No there isn’t enough water, 975 divided by 70 is 13.92 indicating there would only be enough water for 13 experiments and not 15
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3 years ago
Find the value of x in each case. Give reasons to justify your solutions! NEEDED ASAP
Mashcka [7]

Answer:

<em>( x = 17° )</em>

Step-by-step explanation:

Consider the following steps;

m< ZYW = 2x - Given,\\m< YXW = ( 3x - 5 ) - Given,\\m< XYW = 90 - Given,\\\\m< YXW + m< XYW + m< YWX = 180 - Sum of Angles in Triangle,\\( 3x - 5 ) + 90 + m< YWX = 180,\\3x - 5 + 90 + m< YWX = 180,\\3x - 5 + m< YWX = 90,\\m< YWX = 90 - 3x - 5,\\m< YWX = 85 - 3x,\\\\m< ZYW = m< YWX - Alternate Interior Angles,\\2x = 85 - 3x,\\5x = 85,\\Conclusion ; ( x = 17 degrees )

<em>Solution ; ( x = 17° )</em>

4 0
3 years ago
What is the area of this tile?​
yaroslaw [1]

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Step-by-step explanation:

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6 0
3 years ago
Read 2 more answers
Find an equation of the tangent to the curve at the given point by both eliminating the parameter and without eliminating the pa
konstantin123 [22]

Answer:

y=2x-8

Step-by-step explanation:

The given parametric equation is;

x=6+ln(t),y=t^2+3

<h3><u>BY ELIMINATING THE PARAMETER</u></h3>

To eliminate the parameter we make t the subject in one equation and put it inside the other.

We make t the subject in x=6+ln(t) because it is easier.

\Rightarrow x-6=ln(t)

\Rightarrow {e}^{x-6}=e^{ln(t)}

\Rightarrow {e}^{x-6}=t

Or

t={e}^{x-6}

We now substitute this into y=t^2+3.

This gives us;

y=(e^{x-6})^2+3.

\Rightarrow y=e^{2(x-6)}+3.

We have now eliminated the parameter.

The equation of the tangent at (6,4) is given by;

y-y_1=m(x-x_1)

where the gradient function is given by;

\frac{dy}{dx}=2e^{2(x-6)}

We substitute x=6 into the gradient function to obtain the gradient.

\Rightarrow m=2e^{2(6-6)}

\Rightarrow m=2e^0

\Rightarrow m=2

The equation of the tangent becomes

y-4=2(x-6)

We simplify to obtain

y=2x-12+4

y=2x-8

<h3><u>WITHOUT ELIMINATING THE PARAMETER</u></h3>

The given parametric equation is;

x=6+ln(t),y=t^2+3

For x=6+ln(t)

\frac{dx}{dt}=\frac{1}{t}

For y=t^2+3

\frac{dy}{dt}=2t

The slope is given by;

\frac{dy}{dx}=\frac{\frac{dy}{dt} }{\frac{dx}{dt} }

\frac{dy}{dx}=\frac{2t }{\frac{1}{t} }

\frac{dy}{dx}=2t^2

At the point, (6,4), we plug in any of the values into the parametric equation and find the corresponding value for t.

Notice that

When x=6, 6=6+\ln(t)

6-6=\ln(t)

0=\ln(t)

e^0=e^\ln(t)

1=t

when y=4, 4=t^2+3

4-3=t^2

1=t^2

t=\pm1

But the slope is the same when we plug in any of these values for t.

\frac{dy}{dx}=2(\pm1)^2=2

The equation of the tangent becomes

y-4=2(x-6)

We simplify to obtain

y=2x-12+4

y=2x-8

7 0
3 years ago
How much is 7.7 in minutes please it’s a test I need help someone answer
ladessa [460]

Answer:

462 minutes

Step-by-step explanation:

3 0
3 years ago
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