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3 years ago

6

Suppose $a$ and $b$ are the solutions to the quadratic equation $2x^2-3x-6=0$. Find the value of $(a+2)(b+2)$.

1 answer:

Charra [1.4K]3 years ago

4 0

We have that

2x²<span>-3x-6=0
using a graph tool----------> to resolve the second order equation
see the attached figure
the solution is
x=-1.137
x=2.637

therefore
a=-1.137
b=2.637
so
</span>(a+2)*(b+2)-----------> (-1.137+2)*(2.637+2)--------> 0.863*4.637---> 4

the answer is 4

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Solve the triangle. Round your answers to the nearest tenth. Law of sines 6 A. M∠C=34 m ∠ C = 34 , b=25 b = 25 , c=16 c = 16 B.

PolarNik [594]

Answer:

$A = 43.0^o$

$B = 55.0^o$

$BC = 20.0$

Step-by-step explanation:

I will solve this question with the attached triangle

From the attachment, we have:

$\angle C =82^o$

$AB = 29$

$AC = 24$

Required

Solve the triangle

First, we calculate $\angle B$ using sine law:

$\frac{AC}{\sin(B)} = \frac{AB}{\sin(C)}$

This gives:

$\frac{24}{\sin(B)} = \frac{29}{\sin(82^o)}$

$\frac{24}{\sin(B)} = \frac{29}{0.9903}$

Cross multiply

$\sin(B) * 29 = 24 * 0.9903$

$\sin(B) * 29 = 23.7672$

Divide both sides by 29

$\sin(B) = 0.8196$

Take arcsin of both sides

$B = \sin^{-1}(0.8196)$

$B = 55.0^o$

Next, calculate $\angle A$ using:

$A + B + C = 180^o$ --- angles in a triangle

$A + 55^o + 82^o = 180^o$

Collect like terms

$A = 180^o-55.0^o - 82^o$

$A = 43.0^o$

Next, calculate BC using sine laws

$\frac{BC}{\sin(A)} = \frac{AB}{\sin(C)}$

This gives:

$\frac{BC}{\sin(43.0)} = \frac{29}{\sin(82)}$

$\frac{BC}{0.6820} = \frac{29}{0.9903}$

Make BC the subject

$BC = \frac{29*0.6820}{0.9903}$

$BC = 20.0$

6 0

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Bad White [126]

Answer:

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Step-by-step explanation:

by diving 8100 by 15, you get a quotient of 540

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3 years ago

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Drupady [299]

Answer:

x=25

Step-by-step explanation:

60+70=130

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Read 2 more answers

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inysia [295]

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Ludmilka [50]

Answer:

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cross multiply, it'll be

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8 0

2 years ago