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3 years ago

3x+2y=3y line 1 x+y=10 line 2 solve for x/y

Mathematics

1 answer:

Mazyrski [523]3 years ago

3 0

Answer:

x/y=1/3

Step-by-step explanation:

3x+2y=3y

x+y=10

---------------

3x=3y-2y

3x=y

x+3x=10

4x=10

x=10/4=5/2

5/2+y=10

y=10-5/2=20/2-5/2=15/2

x/y=(5/2)/(15/2)=(5/2)(2/15)=5/15=1/3

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A collection of quarters, dimes, and nickels is worth $13.30. If the ratio of quarters to

RideAnS [48]

Answer:

105

Step-by-step explanation:

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2 years ago

Use the discriminant of the following quadratic equation to state the number and type of solutions. 3x^(2)+4x-5=3x+2

krek1111 [17]

3x^(2)+4x-3x-5-2=0
3x^(2)+x-7=0
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2 years ago

Help- and also show the work please-

svetlana [45]

A fraction is a way to describe a part of a whole. The correct option is C.

<h3>What is a Fraction?</h3>

A fraction is a way to describe a part of a whole. such as the fraction ¼ can be described as 0.25.

A proper fraction is a fraction whose numerator is smaller than the denominator. When a proper fraction is raised to a positive exponent greater than 1, then the product decreases but does not reach 0.

Hence, the correct option is C.

Learn more about Fraction:

brainly.com/question/1301963

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1 year ago

Consider the rectangle enlargement. Pls answer the three questions I will mark brainliest!!

Alenkasestr [34]

Answer:

Perimeter of the original Rectangle is

2(4.5+4.3)=2×8.8= <u>17.6</u><u> </u><u>mm</u>

Side length of the enlarged rectangle is

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7 0

2 years ago

Can someone please explain this to me? Thanks!

Makovka662 [10]

Answer: Choice D

$\displaystyle F\ '(x) = 2x\sqrt{1+x^6}\\\\$

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Explanation:

Let g(t) be the antiderivative of $g'(t) = \sqrt{1+t^3}$ . We don't need to find out what g(t) is exactly.

Recall by the fundamental theorem of calculus, we can say the following:

$\displaystyle \int_{a}^{b} g'(t)dt = g(b)-g(a)$

This theorem ties together the concepts of integrals and derivatives to show that they are basically inverse operations (more or less).

So,

$\displaystyle F(x) = \int_{\pi}^{x^2}\sqrt{1+t^3}dt\\\\ \displaystyle F(x) = \int_{\pi}^{x^2}g'(t)dt\\\\ \displaystyle F(x) = g(x^2) - g(\pi)\\\\$

From here, we apply the derivative with respect to x to both sides. Note that the $g(\pi)$ portion is a constant, so $g'(\pi) = 0$

$\displaystyle F(x) = g(x^2) - g(\pi)\\\\ \displaystyle F \ '(x) = \frac{d}{dx}[g(x^2)-g(\pi)]\\\\\displaystyle F\ '(x) = \frac{d}{dx}[g(x^2)] - \frac{d}{dx}[g(\pi)]\\\\ \displaystyle F\ '(x) = \frac{d}{dx}[x^2]*g'(x^2) - g'(\pi) \ \text{ .... chain rule}\\\\$

$\displaystyle F\ '(x) = 2x*g'(x^2) - 0\\\\ \displaystyle F\ '(x) = 2x*g'(x^2)\\\\ \displaystyle F\ '(x) = 2x\sqrt{1+(x^2)^3}\\\\ \displaystyle F\ '(x) = \boldsymbol{2x\sqrt{1+x^6}}\\\\$

Answer is choice D

5 0

2 years ago

Read 2 more answers