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3 years ago

Can someone help me with this math homework please!

Mathematics

1 answer:

Daniel [21]3 years ago

8 0

Answer:

1st, 2nd and 6th options

Step-by-step explanation:

Given

$\frac{2}{3}$ - x + $\frac{1}{6}$ = 6x ( multiply through by 6 to clear the fractions )

4 - 6x + 1 = 36x ← option 1 will have the same solution set

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Adding the 2 fractions on the left side gives

$\frac{4}{6}$ + $\frac{1}{6}$ - x = 6x , that is

$\frac{5}{6}$ - x = 6x ← option 2 will have the same solution set

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

From

4 - 6x + 1 = 36x ( add 6x to both sides )

5 = 42x ← option 6 will have the same solution set

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Evaluate the degree of the polynomial (y3 – 2) ( y2 + 11), expand using algebraic expressions and answer.

posledela

Answer:

y^5+11y^3-2y^3-22

Step-by-step explanation:

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4 years ago

1. Paula has x cups of food in a container to feed her dogs. She pours 1.5 cups of food into their bowls. There is now 5.25 cups

kifflom [539]

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3 years ago

Solve 5y'' + 3y' – 2y = 0, y(0) = 0, y'(0) = 2.8 y(t) = 0 Preview

mario62 [17]

Answer: The required solution is

$y(t)=-\dfrac{7}{3}e^{-t}+\dfrac{7}{3}e^{\frac{1}{5}t}.$

Step-by-step explanation: We are given to solve the following differential equation :

$5y^{\prime\prime}+3y^\prime-2y=0,~~~~~~~y(0)=0,~~y^\prime(0)=2.8~~~~~~~~~~~~~~~~~~~~~~~~(i)$

Let us consider that

$y=e^{mt}$ be an auxiliary solution of equation (i).

Then, we have

$y^prime=me^{mt},~~~~~y^{\prime\prime}=m^2e^{mt}.$

Substituting these values in equation (i), we get

$5m^2e^{mt}+3me^{mt}-2e^{mt}=0\\\\\Rightarrow (5m^2+3y-2)e^{mt}=0\\\\\Rightarrow 5m^2+3m-2=0~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~[\textup{since }e^{mt}\neq0]\\\\\Rightarrow 5m^2+5m-2m-2=0\\\\\Rightarrow 5m(m+1)-2(m+1)=0\\\\\Rightarrow (m+1)(5m-1)=0\\\\\Rightarrow m+1=0,~~~~~5m-1=0\\\\\Rightarrow m=-1,~\dfrac{1}{5}.$

So, the general solution of the given equation is

$y(t)=Ae^{-t}+Be^{\frac{1}{5}t}.$

Differentiating with respect to t, we get

$y^\prime(t)=-Ae^{-t}+\dfrac{B}{5}e^{\frac{1}{5}t}.$

According to the given conditions, we have

$y(0)=0\\\\\Rightarrow A+B=0\\\\\Rightarrow B=-A~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~(ii)$

and

$y^\prime(0)=2.8\\\\\Rightarrow -A+\dfrac{B}{5}=2.8\\\\\Rightarrow -5A+B=14\\\\\Rightarrow -5A-A=14~~~~~~~~~~~~~~~~~~~~~~~~~~~[\textup{Uisng equation (ii)}]\\\\\Rightarrow -6A=14\\\\\Rightarrow A=-\dfrac{14}{6}\\\\\Rightarrow A=-\dfrac{7}{3}.$

From equation (ii), we get

$B=\dfrac{7}{3}.$

Thus, the required solution is

$y(t)=-\dfrac{7}{3}e^{-t}+\dfrac{7}{3}e^{\frac{1}{5}t}.$

7 0

3 years ago

PLEASE ALSO EXPLAIN HOW TO SOLVE THIS. 20 POINTS

eimsori [14]

B.) Hibiscus Street

The key thing to remember is the slope of the lines. The equation for a line in slope intercept form is
y = ax + b
where
a = slope
b = y intercept

So the equation for Oak Street is y = 2/3x - 7. So it's slope is 2/3. And any street that has the same slope will be given a tree name. And any street that's perpendicular will be given a flower name. You can determine is a line is perpendicular if it has a slope that's the negative reciprocal. So a street that's perpendicular to Oak street will have a slope of -3/2.

Now you've just been given the equation to a new street that's y = -3/2x - 2. Since the slope is -3/2 and Oak street has a slope of 2/3, the new street is perpendicular to Oak street. And given the naming scheme, that means that the new street will have the name of a flower. So let's look at the available street names and pick a flower.
<span>
A.) Weeping Willow Street
* Nope a Weeping Willow is a tree. So this name won't work.

B.) Hibiscus Street
* Yes. A Hibiscus is a flower, so this name is suitable.

C.) Oak Street
* Nope. Not only is this a tree instead of a flower, but there's already an Oak street. So bad choice.

D.) Panther Street
* Nope, this is an animal, not a flower. Bad choice.

</span>

6 0

3 years ago

URGENT!!! SOLVE BY TAKING THE SQUARE ROOTS. 50 POINTS

il63 [147K]

Answer:

Step-by-step explanation:

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3 years ago