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2 years ago

14

The archway of the main entrance of a university is modeled by the quadratic equation y = -x2 + 6x. The university is hanging a

banner at the main entrance at an angle defined by the equation 4y = 21 − x. At what points should the banner be attached to the archway?
A(1, 5.5) and (5.25, 6.56)
B(1, 5) and (5.25, 3.94)
C(1.5, 4.87) and (3.5, 4.37)
D(1.5, 5.62) and (3.5, 6.12)
EThere is no real solution.

2 answers:

katrin [286]2 years ago

5 0

Basically look for the intersection points of these two functions by setting them equal to each other.

-x^2 + 6x = 21/4 - x/4

-x^2 + (25/4)x - 21/4 = 0

x = 1 and x = 21/4

Plug these points back to any of the two original equations, you will get

(1,5) and (21/4, 63/16)

denpristay [2]2 years ago

4 0

1.5 ........ 5.25..... 3.94

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An experiment was performed to compare the abrasive wear of two different laminated materials. Twelve pieces of material 1 were

vladimir1956 [14]

Answer:

The calculated value Z= 4.8389> 1.96 at 0.05 level of significance.

The null hypothesis is rejected.

There is significance difference between that the abrasive wear of material 1 not exceeds that of material 2 by more than 2 units

Step-by-step explanation:

<u>Step:-(1)</u>

Given data the samples of material 1 gave an average (coded) wear of 85 units with a sample standard deviation of 4

Mean of the first sample x₁⁻ =85

standard deviation of the first sample S₁ = 4

Given data the samples of material 2 gave an average of 81 and a sample standard deviation of 5.

Mean of the first sample x₂⁻ =81

standard deviation of the first sample S₂ = 5

<u>Step :-2</u>

<u>Null hypothesis: H₀:</u> there is no significance difference between that the abrasive wear of material 1 exceeds that of material 2 by more than 2 units

<u>Alternative hypothesis :H₁: </u>there is significance difference between that the abrasive wear of material 1 exceeds that of material 2 by more than 2 units

Assume the populations to be approximately normal with equal variances.σ₁² =σ₂²

The test statistic

$Z= \frac{x_{1} -x_{2} }{\sqrt{\frac{S^2_{1} }{n_{1} } +\frac{S^2_{2} }{n_{2} } } }$

Given n₁=n₂=60.

$Z= \frac{85-81 }{\sqrt{\frac{(4)^2 }{60 } +\frac{5^2 }{60} } }$

On calculation, we get

Z = $\frac{4}{\sqrt{0.6833} }$

z = 4.8389

The tabulated value Z =1.96 at 0.05 level of significance.

The calculated value Z= 4.8389> 1.96 at 0.05 level of significance.

The null hypothesis is rejected.

Conclusion:-

there is significance difference between that the abrasive wear of material 1 not exceeds that of material 2 by more than 2 units.

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2 years ago

Marty is asked to draw triangles with side lengths of 4 units and 2 units, and a non-included angle of 30°. Select all the trian

777dan777 [17]

Answer:

The drawn in the attached figure

see the explanation

Step-by-step explanation:

<em>First case</em>

In the triangle ABC

Let

$a=4\ units\\b=2/ units\\B=30^o$

Applying the law of sines

Find the measure of angle A

$\frac{a}{sin(A)}=\frac{b}{sin(B)}$

substitute the given values

$\frac{4}{sin(A)}=\frac{2}{sin(30^o)}$

$sin(A)=1$

so

$A=90^o$

Find the measure of angle C

In a right triangle

we know that

$B+C=90^o$ ----> by complementary angles

$B=30^o$

therefore

$C=60^o$

Find the length side c

Applying the law of sines

$\frac{c}{sin(C)}=\frac{b}{sin(B)}$

substitute the given values

$\frac{c}{sin(60^o)}=\frac{2}{sin(30^o)}$

$c=2\sqrt{3}\ units$

therefore

The dimensions of the triangle are

$A=90^o$

$B=30^o$

$C=60^o$

$a=4\ units\\b=2\ units\\c=2\sqrt{3}=3.46\ units$

<em>Second case</em>

In the triangle ABC

Let

$a=4\ units\\b=2/ units\\A=30^o$

Applying the law of sines

Find the measure of angle B

$\frac{a}{sin(A)}=\frac{b}{sin(B)}$

substitute the given values

$\frac{4}{sin(30^o)}=\frac{2}{sin(B)}$

$sin(B)=0.25$

so

using a calculator

$B=14.48^o$

Find the measure of angle C

we know that

The sum of the interior angles in any triangle must be equal to 180 degrees

so

$A+B+C=180^o$

$A=30^o\\B=14.48^o$

therefore

$30^o+14.48^o+C=180^o$

$C=135.52^o$

Find the length side c

Applying the law of sines

$\frac{c}{sin(C)}=\frac{a}{sin(A)}$

substitute the given values

$\frac{c}{sin(135.52^o)}=\frac{4}{sin(30^o)}$

$c=5.61\ units$

therefore

The dimensions of the triangle are

$A=30^o$

$B=14.48^o$

$C=135.52^o$

$a=4\ units\\b=2\ units\\c=5.61\ units$

see the attached figure to better understand the problem

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ExtremeBDS [4]

Answer:

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Step-by-step explanation:

To simplify:

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First of all, let us write 18 and 162 as product of prime factors:

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