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Anna11 [10]
2 years ago
9

PQ⊥PS , m∠QPR=7x−9, m∠RPS=4x+22 Find : m∠QPR

Mathematics
1 answer:
emmainna [20.7K]2 years ago
3 0

Given:

It is given that,

PQ ⊥ PS and

∠QPR = 7x-9

∠RPS = 4x+22

To find the value of ∠QPR.

Formula

As per the given problem PR lies between PQ and PS,

So,

∠QPR+∠RPS = 90°

Now,

Putting the values of ∠RPS and ∠QPR we get,

7x-9+4x+22 = 90

or, 11x = 90-22+9

or, 11x = 77

or, x = \frac{77}{11}

or, x = 7

Substituting the value of x = 7 in ∠QPR we get,

∠QPR = 7(7)-9

or, ∠QPR = 40^\circ

Hence,

The value of ∠QPR is 40°.

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Use the position function s(t) = −16t² + 400, which gives the height (in feet) of an object that has fallen for t seconds from a
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Answer:

160m/s

Step-by-step explanation:

The object can hit the ground when t = a; meaning that s(a) = s(t) = 0

So, 0 = -16a² + 400

16a² = 400

a² = 25

a = √25

a = 5 (positive 5 only because that's the only physical solution)

The instantaneous velocity is

v(a) = lim(t->a) [s(t) - s(a)]/[t-a)

Where s(t) = -16t² + 400

and s(a) = -16a² + 400

v(a) = Lim(t->a) [-16t² + 400 + 16a² - 400]/(t-a)

v(a) = Lim(t->a) (-16t² + 16a²)/(t-a)

v(a) = lim (t->a) -16(t² - a²)(t-a)

v(a) = -16lim t->a (t²-a²)(t-a)

v(a) = -16lim t->a (t-a)(t+a)/(t-a)

v(a) = -16lim t->a (t+a)

But a = t

So, we have

v(a) = -16lim t->a 2a

v(a) = -32lim t->a (a)

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7 0
1 year ago
A rectangular portrait measures 50cm by 70cm. It is surrounded by a rectangular frame of uniform width. If the area of the frame
snow_tiger [21]

Let us assume uniform width = x cm wide.

Length of rectangular portrait = 50cm and width of rectangular portrait = 70cm.

Therefore,  length of rectangle made by frame = 50 + x+x = (2x+50) cm.

And width of rectangle made by frame = 70+x+x = (2x+70) cm.

We know, the area of rectangular portrait = 50 × 70 = 3500 cm^2.

Total area of the rectangle made by frame would be =  (2x+50) * (2x+70)

We know,

Actual area of frame = Area of rectangle made by frame -  area of rectangular portrait.

We also given "the area of the frame is the same as the area of the portrait."

We can setup an equation now,

 3500 = (2x+50) * (2x+70) - 3500.

Subtracting 3500 from both sides, we get

3500-3500 = (2x+50) * (2x+70) - 3500-3500.

0 = (2x+50) * (2x+70) -7000.

FOIL (2x+50) * (2x+70), we get

0 = 2x*2x +2x*70 + 50*2x +50*70 - 7000.

0 = 4x^2 +140x +100x +3500 -7000.

4x^2 +240x -3500 = 0.

Dividing whole equation by 4, we get

x^2 +60x - 875 =0

Applying quadratic formula =\frac{-b\pm \sqrt{b^2-4ac}}{2a}, we get

=\frac{-60\pm \sqrt{60^2-4\cdot \:1\left(-875\right)}}{2\cdot \:1}

x=\frac{-60+\sqrt{60^2-4\cdot \:1\left(-875\right)}}{2\cdot \:1}=5\left(\sqrt{71}-6\right)

x=\frac{-60-\sqrt{60^2-4\cdot \:1\left(-875\right)}}{2\cdot \:1}:\quad -5\left(6+\sqrt{71}\right)

We cant take negative value.

So, x=5\left(\sqrt{71}-6\right)=12.13

We could take approximately 12 cm.



7 0
2 years ago
To evaluate the effect of a treatment, a sample of n=8 is obtained from a population with a mean of μ=40 , and the treatment is
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Answer:

Step-by-step explanation:

a)

Test statistic:

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t=-2.5

df=8-1=7

critical, t ,values=+/-2.365

here test statistic lie in rejection region,that why null hypothesis fails  

so Yes, its significant.

b)

Test statistic:

t=\frac{35-40}{\sqrt{\frac{72}{8} } }

t=-1.67

df=8-1=7

critical, t ,values=+/-2.365

c)

sample variability increases, therefore likelihood of rejecting the null hypothesis decreases.

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