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3 years ago

Put the following into slope intercept form. -5x + 3y = 12

Mathematics

1 answer:

polet [3.4K]3 years ago

5 0

Answer:

y=5/3x+4

Step-by-step explanation:

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a line has the equation 4y=3x-8 which is the equation of a line parallel to this line that has a y intercpt of 4?

Nikolay [14]

Answer:

y=34x+2.625

y=34x+2.75

y=−34x+1.25

y=34x+1.25

y=−34x+2.75

Correct answer:

y=−34x+2.75

Step-by-step explanation:

5 0

3 years ago

Melia put the $30 her Grandmother gave her for her tenth birthday into her piggy bank. Each week after

Vladimir [108]

This is a sequence because the points form a straight line. The expression is: t(n) = 30 + 2x, x is number of weeks. On her 18th birthday, that is 8 years after her 10th. that means 416 weeks have passed since her 10th birthday. Plug 416 into the equation to get: t(416) = 30 + 2(416) = $862. She will have $862 to Spend.

3 0

3 years ago

How do I do this help me please

Dmitrij [34]

Answer:

The answer is 3x.

Step-by-step explanation:

I broke it down on paper.

3 0

3 years ago

One of Julian's friends also has a rectangular aquarium. The dimensions of his friend's tank are each exactly 1 1/4 times the di

daser333 [38]

Answer:

468.75 square inches

Step-by-step explanation:

Let the length of Julian's tank be x

Let the width of Julian's tank be y

So, Area of Julian's tank= $Length \times Width =xy$

Since we are given that Julian's tank is approximately 300 square inches

So, $xy=300$ --1

Now we are given that The dimensions of his friend's tank are each exactly $1\frac{1}{4} =\frac{5}{4}$ times the dimensions of Julian's tank.

So,length of his friend's tank = $\frac{5}{4}x$

So,width of his friend's tank = $\frac{5}{4}y$

So, Area of his friend's tank= $\frac{5}{4}x \times \frac{5}{4}y$

= $\frac{25}{16}xy$

= $\frac{25}{16}(300)$ (Using 1)

= $468.75$

Hence the approximate area of the rectangular base of his friend's tank is 468.75 square inches.

So, Option D is correct.

8 0

3 years ago

Calculus help please!!! Click the link to see the integral.

iren [92.7K]

Noice

ook, so use u subsitution

remember that dy/dx(sin(x))=cos(x)

u=sin(x)
du/dx=cos(x)
du=cos(x) dx
so then

$\int\limits{ \frac{cos(x)}{ \sqrt{sin(x)} } } \, dx$
replace
$\int\limits{ \frac{1}{ \sqrt{u} } } \, du$
we do some stuff
$\int\limits{ u^ \frac{-1}{2} } } \, du$
$\frac{u^ \frac{1}{2} }{ \frac{1}{2} }$
$2 \sqrt{u}$
u=sin(x)
$2 \sqrt{sin(x)}$

3 0

3 years ago