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3 years ago

6

In the graph, the yellow highlighted part represents the feasible region of all possible solutions to what kind of solution meth

od?
A. Linear equations
B. Linear functions
C. Linear lines
D. Linear programming

1 answer:

julia-pushkina [17]3 years ago

3 0

I think the answer is A

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Increase £16470.45 by 3.5%

Korvikt [17]

Try this:
1. if to increase by 3.5%, then 16470.45 is 100% and new number is 103.5%; it means 1.035 times;
2. according to the previous issue result is 16470.45*1.035=17046.91575.
answer: 17046.91575

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3 years ago

Find the square root of 144

Gennadij [26K]

Answer:

12

Step-by-step explanation:

Hi there !

144 | 2

72 | 2

36 | 2

18 | 2

9 | 3

3 | 3

1

144 = 2⁴ × 3²

√144 = √2⁴×3² = 2²×3 = 4×3 = 12

Good luck !

8 0

3 years ago

Find the exact value of the expression.<br> tan( sin−1 (2/3)− cos−1(1/7))

Sonja [21]

Answer:

$\tan(a-b)=\frac{2\sqrt{5}-20\sqrt{3}}{5+8\sqrt{15}}$

Step-by-step explanation:

I'm going to use the following identity to help with the difference inside the tangent function there:

$\tan(a-b)=\frac{\tan(a)-\tan(b)}{1+\tan(a)\tan(b)}$

Let $a=\sin^{-1}(\frac{2}{3})$ .

With some restriction on $a$ this means:

$\sin(a)=\frac{2}{3}$

We need to find $\tan(a)$ .

$\sin^2(a)+\cos^2(a)=1$ is a Pythagorean Identity I will use to find the cosine value and then I will use that the tangent function is the ratio of sine to cosine.

$(\frac{2}{3})^2+\cos^2(a)=1$

$\frac{4}{9}+\cos^2(a)=1$

Subtract 4/9 on both sides:

$\cos^2(a)=\frac{5}{9}$

Take the square root of both sides:

$\cos(a)=\pm \sqrt{\frac{5}{9}}$

$\cos(a)=\pm \frac{\sqrt{5}}{3}$

The cosine value is positive because $a$ is a number between $-\frac{\pi}{2}$ and $\frac{\pi}{2}$ because that is the restriction on sine inverse.

So we have $\cos(a)=\frac{\sqrt{5}}{3}$ .

This means that $\tan(a)=\frac{\frac{2}{3}}{\frac{\sqrt{5}}{3}}$ .

Multiplying numerator and denominator by 3 gives us:

$\tan(a)=\frac{2}{\sqrt{5}}$

Rationalizing the denominator by multiplying top and bottom by square root of 5 gives us:

$\tan(a)=\frac{2\sqrt{5}}{5}$

Let's continue on to letting $b=\cos^{-1}(\frac{1}{7})$ .

Let's go ahead and say what the restrictions on $b$ are.

$b$ is a number in between 0 and $\pi$ .

So anyways $b=\cos^{-1}(\frac{1}{7})$ implies $\cos(b)=\frac{1}{7}$ .

Let's use the Pythagorean Identity again I mentioned from before to find the sine value of $b$ .

$\cos^2(b)+\sin^2(b)=1$

$(\frac{1}{7})^2+\sin^2(b)=1$

$\frac{1}{49}+\sin^2(b)=1$

Subtract 1/49 on both sides:

$\sin^2(b)=\frac{48}{49}$

Take the square root of both sides:

$\sin(b)=\pm \sqrt{\frac{48}{49}$

$\sin(b)=\pm \frac{\sqrt{48}}{7}$

$\sin(b)=\pm \frac{\sqrt{16}\sqrt{3}}{7}$

$\sin(b)=\pm \frac{4\sqrt{3}}{7}$

So since $b$ is a number between $0$ and $\pi$ , then sine of this value is positive.

This implies:

$\sin(b)=\frac{4\sqrt{3}}{7}$

So $\tan(b)=\frac{\sin(b)}{\cos(b)}=\frac{\frac{4\sqrt{3}}{7}}{\frac{1}{7}}$ .

Multiplying both top and bottom by 7 gives:

$\frac{4\sqrt{3}}{1}= 4\sqrt{3}$ .

Let's put everything back into the first mentioned identity.

$\tan(a-b)=\frac{\tan(a)-\tan(b)}{1+\tan(a)\tan(b)}$

$\tan(a-b)=\frac{\frac{2\sqrt{5}}{5}-4\sqrt{3}}{1+\frac{2\sqrt{5}}{5}\cdot 4\sqrt{3}}$

Let's clear the mini-fractions by multiply top and bottom by the least common multiple of the denominators of these mini-fractions. That is, we are multiplying top and bottom by 5:

$\tan(a-b)=\frac{2 \sqrt{5}-20\sqrt{3}}{5+2\sqrt{5}\cdot 4\sqrt{3}}$

$\tan(a-b)=\frac{2\sqrt{5}-20\sqrt{3}}{5+8\sqrt{15}}$

4 0

3 years ago

Plz help me answer this

timofeeve [1]

Answer:

(-2, -8)

x = -2

y = -8

General Formulas and Concepts:

<u>Pre-Algebra</u>

Order of Operations: BPEMDAS
Equality Properties

<u>Algebra I</u>

Solving systems of equations using substitution/elimination

Step-by-step explanation:

<u>Step 1: Define systems</u>

13x - 6y = 22

x = y + 6

<u>Step 2: Solve for </u><em><u>y</u></em>

<em>Substitution</em>

Substitute in <em>x</em>: 13(y + 6) - 6y = 22
Distribute 13: 13y + 78 - 6y = 22
Combine like terms: 7y + 78 = 22
Isolate <em>y</em> term: 7y = -56
Isolate <em>y</em>: y = -8

<u>Step 3: Solve for </u><em><u>x</u></em>

Define original equation: x = y + 6
Substitute in <em>y</em>: x = -8 + 6
Add: x = -2

4 0

3 years ago

The formula for wind chill (in degrees Fahrenheit) is given byC = 35.74 + 0.6215T ? 35.75v0.16 + 0.4275Tv0.16 where is the wind

PIT_PIT [208]

Answer:

dC =0.6899T + 30.02

dC/C = (0.6899T + 30.02)/(35.74 + 0.6215T - 35.75V^0.16 + 0.4275Tv^0.16

Step-by-step explanation:

C= 35.74 + 0.6215T - 35.75v^0.16 + 0.4275Tv^0.16

dC = 35.74 + 0.6215T - (35.75×0.16) +:(0.4275×0.16)

dC = 35.74 + 0.6215T - 5.72 + 0.684T

dC = 0.6899T + 30.02

dC/C = (0.6899 + 30.02)/(35.74 + 0.6215T - 35.75Tv^0.16 + 0.4275Tv^0.16

4 0

4 years ago