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patriot [66]
3 years ago
7

Evan plant grass in a rectangular space behind the clubhouse

Mathematics
1 answer:
Nastasia [14]3 years ago
6 0
A= l x b
70 = 14b
divide both sides by 14
b= 5
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Find the tangent line equation of the curve at the given point. Y=arcsin(7x) at the point where x=sqrt2/14
Mumz [18]

Answer:

Y-\frac{\pi}{2} =\frac{\pi}{2} (x+\sqrt{\frac{1}{7}})

Step-by-step explanation:

The equation of the curve is

Y = sin^{-1}(7x)

To find the equation of tangent we need to differentiate this equation w.r.t x

So, differentiating we get

Y'=\frac{7}{\sqrt{1-49x^2} }

This would give the slope of the tangent line at any given point of which x coordinate is known. In the present case it is  x = \sqrt{\frac{1}{7} }

Then slope would accordingly be

Y'=\frac{7}{\sqrt{1-49/49} }

= ∞

For, x = \sqrt{\frac{1}{7} }, Y = sin^{-1}(7/7)= \pi/2

Equation of tangent line, in the point slope form, would be Y-\frac{\pi}{2} =\frac{\pi}{2} (x+\sqrt{\frac{1}{7}})

4 0
3 years ago
Complete the square to make a perfect square trinomial. Then, write the result as a binomial squared. q2+11q
Sedbober [7]

Answer:

(q+\frac{11}{2})^2-\frac{121}{4}

Step-by-step explanation:

We have been given an expression q^2+11q. We are asked to complete the square to make a perfect square trinomial. Then, write the result as a binomial squared.

We know that a perfect square trinomial is in form a^2+2ab+b^2.

To convert our given expression into perfect square trinomial, we need to add and subtract (\frac{b}{2})^2 from our given expression.

We can see that value of b is 11, so we need to add and subtract (\frac{11}{2})^2 to our expression as:

q^2+11q+(\frac{11}{2})^2-(\frac{11}{2})^2

Upon comparing our expression with (a+b)^2=a^2+2ab+b^2, we can see that a=q, 2ab=11q and b=\frac{11}{2}.

Upon simplifying our expression, we will get:

(q+\frac{11}{2})^2-\frac{11^2}{2^2}

(q+\frac{11}{2})^2-\frac{121}{4}

Therefore, our perfect square would be (q+\frac{11}{2})^2-\frac{121}{4}.

8 0
3 years ago
Read 2 more answers
Which of the following could be the equation of a line parallel to the line y=-4x+1
saveliy_v [14]
It would be D.
Because the slope is the same but the yint is different
5 0
3 years ago
Read 2 more answers
2. The acute angle below has a degree measurement that is one fifth of the degree measure of the obtuse
Ksenya-84 [330]

Answer:

Acute angle = 30°

Obtuse angle = 150°

Step-by-step explanation:

Method 1:

Let x represent the measurement of the obtuse angle

Obtuse angle = x

Acute angle = ⅕ of x = x/5

Thus:

x + x/5 = 180° (angels on a straight line)

Solve for x

(5x + x)/5 = 180

Multiply both sides by 5

5x + x = 180 × 5

6x = 900

x = 900/6

x = 150

Obtuse angle = 150°

Acute angle = x/5 = 150/5 = 30°

Method 2:

Since acute angle = ⅕ of the obtuse angle, therefore,

Obtuse angle = 5*acute angle

Let acute angle = x

Obtuse angle = 5x

Equation:

5x + x = 180° (angles on a straight line)

Solve for x

6x = 180

x = 180/6

x = 30

Acute angle = x = 30°

Obtuse angle = 5x = 5*30 = 150°

8 0
3 years ago
Evaluate the expression below when y = -3.
Rudik [331]
Start with the given equation:
2y+7
Substitute known values:
2(-3)+7
Evaluate:
-6+7=1


Similarly:
-4y-10
-4(-3)-10
12-10=2
8 0
3 years ago
Read 2 more answers
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