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3 years ago

The following system of equations is

Mathematics

1 answer:

Snezhnost [94]3 years ago

7 0

Answer:

equivalent

Step-by-step explanation:

when you divide the second one by 2 you'll get the first one

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Given: ABCD is a rhombus, <br> Perimeter = 20,<br> AC:BD = 4:3 Find:<br> Area of ABCD.

REY [17]

4a=20⇒a=5

we know diagonals are perpendicular. if u look at one of the right triangles, you can see that for hypotenuse to be 5, the legs have to be 4 and 3 to satisfy the given ratio 4:3.

so, diagonals are 8 and 6 as we know that diagonals bisect each other.

Area=8*6/2=24

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3 years ago

What is the value represented by the letter A on the box plot of data?

Lady_Fox [76]

Answer: A = 5

The Wise Orange found an image that may be helpful:

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3 years ago

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Helppppppppppppppppppppppppppppp

Vladimir [108]

Answer:

$\sf \dfrac{3}{4y}$

Step-by-step explanation:

Simplify:

45 = 5 * 3 * 3

60 = 5 * 3 * 2 * 2

GCF = 5*3 = 15

$\sf \dfrac{45x^2}{60x^2y}=\dfrac{45}{60y}$

$\sf =\dfrac{45 \div 15}{60y \div 15}\\\\= \dfrac{3}{4y}$

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1 year ago

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Eliza is a chef at a restaurant. she is making spaghetti sauce and put 21 tomatoes and 6 onions in the pot. what is the ratio of

seraphim [82]

These are not hard....u just have to pay very close attention to the wording. Keep in mind ratios can be written as fractions too.

ratio of tomatoes to onions = 21:6 which reduces to 7:2...because there is 21 tomatoes and 6 onions. (21/6 reduces to 7/2...ratios written as fractions)

However, if it would have asked the ratio of onions to tomatoes, it would have been 6:21....which reduces to 2:7....so pay attention to the wording on problems such as these

so ur answer is 7:2

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3 years ago

Please help me for the love of God if i fail I have to repeat the class

Elena-2011 [213]

$\theta$ is in quadrant I, so $\cos\theta>0$ .

$x$ is in quadrant II, so $\sin x>0$ .

Recall that for any angle $\alpha$ ,

$\sin^2\alpha+\cos^2\alpha=1$

Then with the conditions determined above, we get

$\cos\theta=\sqrt{1-\left(\dfrac45\right)^2}=\dfrac35$

and

$\sin x=\sqrt{1-\left(-\dfrac5{13}\right)^2}=\dfrac{12}{13}$

Now recall the compound angle formulas:

$\sin(\alpha\pm\beta)=\sin\alpha\cos\beta\pm\cos\alpha\sin\beta$

$\cos(\alpha\pm\beta)=\cos\alpha\cos\beta\mp\sin\alpha\sin\beta$

$\sin2\alpha=2\sin\alpha\cos\alpha$

$\cos2\alpha=\cos^2\alpha-\sin^2\alpha$

as well as the definition of tangent:

$\tan\alpha=\dfrac{\sin\alpha}{\cos\alpha}$

Then

1. $\sin(\theta+x)=\sin\theta\cos x+\cos\theta\sin x=\dfrac{16}{65}$

2. $\cos(\theta-x)=\cos\theta\cos x+\sin\theta\sin x=\dfrac{33}{65}$

3. $\tan(\theta+x)=\dfrac{\sin(\theta+x)}{\cos(\theta+x)}=-\dfrac{16}{63}$

4. $\sin2\theta=2\sin\theta\cos\theta=\dfrac{24}{25}$

5. $\cos2x=\cos^2x-\sin^2x=-\dfrac{119}{169}$

6. $\tan2\theta=\dfrac{\sin2\theta}{\cos2\theta}=-\dfrac{24}7$

7. A bit more work required here. Recall the half-angle identities:

$\cos^2\dfrac\alpha2=\dfrac{1+\cos\alpha}2$

$\sin^2\dfrac\alpha2=\dfrac{1-\cos\alpha}2$

$\implies\tan^2\dfrac\alpha2=\dfrac{1-\cos\alpha}{1+\cos\alpha}$

Because $x$ is in quadrant II, we know that $\dfrac x2$ is in quadrant I. Specifically, we know $\dfrac\pi2$ , so $\dfrac\pi4$ . In this quadrant, we have $\tan\dfrac x2>0$ , so

$\tan\dfrac x2=\sqrt{\dfrac{1-\cos x}{1+\cos x}}=\dfrac32$

8. $\sin3\theta=\sin(\theta+2\theta)=\dfrac{44}{125}$

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3 years ago