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3 years ago

HELP PLEASE!!!! I NEED IT ASAP!

Mathematics

1 answer:

Vsevolod [243]3 years ago

6 0

Answer:

y = -3 and x = 2

Step-by-step explanation:

Multiply 1st equation by - 3 and add with second equation you will get -y = 3 and you will substitute y to the 1st equation

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3.56 kilograms of pears.

3/5 of 8.9 is 5.34
8.9 - 5.34 = 3.56

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3 years ago

Use the following function rule to find f(6).<br> f(x) = 6 - x<br> f(0) =

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3 years ago

Please help and solve for x! (: thank you in advance.

Vera_Pavlovna [14]

First look at the given numbers, especially 123 degrees which if you add up the other side it equals 180 degrees so:
180-123= 57
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4 years ago

There is a 22% chance it will rain tomorrow and a 6% chance of strong winds. What is the chance it will stay dry

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Step-by-step explanation:

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2 years ago

Q13<br> Please help me solve this….

KonstantinChe [14]

Answer:

Choice C

$\frac{\sqrt{186}+\sqrt{15}}{18}$

Step-by-step explanation:

The quadrant in which an angle lies determines the signs of the trigonometric functions sin, cos and tan

If an angle Θ lies in quadrant IV, cos(Θ) is positive and both sin(Θ) and tan(Θ) are negative

Two of the trigonometric identities we can use are

1. $\sin^2(\theta) + cos^2(\theta) = 1$ and

2. $\cos(A-B) = \cos A\cos B - \sin A\sin B$

Using identity 1, we can solve for cos(s) and cos(t)

$sin(s)=-\frac{\sqrt{3}}{3}, sin^{2}(s)=\frac{3}{9}=\frac{1}{3}\\cos^{2}(s)=1-sin^{2}(s)=1-\frac{1}{3}=\frac{2}{3}; cos(s)=\pm\sqrt{\frac{2}{3}}\\\\$

$sin(t)=-\frac{\sqrt{5}}{6}, sin^{2}(t)=\frac{5}{36}cos^{2}(t)=1-sin^{2}(t)=1-\frac{5}{36}=\frac{31}{36};cos(t)=\pm\frac{\sqrt{31}}{6}$

Since both angles lie in quadrant IV, both cos(s) and cos(t) must be positive so we only consider the positive signs of both values

Using identity 2, we can solve for cos(s-t)

$cos(s-t)=cos(s)cos(t)+sin(s)sin(t)=\frac{\sqrt{2}}{\sqrt{3}}.\frac{\sqrt{31}}{6}+(-\frac{\sqrt{3}}{3})(-\frac{\sqrt{5}}{6})$

Multiplying numerator and denominator of the first term by $\sqrt{3}$ gives us the final expression as

$\frac{\sqrt{3}}{\sqrt{3}}\frac{\sqrt{2}}{\sqrt{3}}.\frac{\sqrt{31}}{6}+(-\frac{\sqrt{3}}{3})(-\frac{\sqrt{5}}{6})=\frac{\sqrt{186}}{18}+\frac{\sqrt{15}}{18}= \frac{\sqrt{186}+\sqrt{15}}{18}$

4 0

2 years ago