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3 years ago

9

jordan want to double the recipe for the cake he decides to use 4 cup of sugar and 7 cups of flour. is jordan correct or incorre

ct? explain your answer

1 answer:

andrew11 [14]3 years ago

3 0

2 1/3 cups of sugar and 3/5 of flour that is the answer

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What is the equation of the line that is parallel to the line y=-1/3x+4 and passes through the point (6,5)?

natulia [17]

Parallel lines have the same slope. The slope of y = -1/3 x + 4 is -1/3.

The equation is

(y - 5) = -1/3 (x - 6)

y = -1/3 x -1/3 (- 6) + 5 = -1/3 x + 2 + 5 = -1/3 + 7

The equation is: y = - 1/3 x + 7

y = -1/3 x

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3 years ago

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PLEASE ANSWER I WILL GIVE BRAINLIEST TO WHOEVER ANSWERS GOOD PLEASE

Scilla [17]

Answer:

Your answer is dilation.

Step-by-step explanation:

A dilation either shrinks or makes the shape bigger. It changes the size.

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3 years ago

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Someone please help me with this asap

Leokris [45]

Answer:

All of these terms are dollar amounts.

Step-by-step explanation:

Here, x + y = 10 (where these numerals represent counts)

(1/2)x + (1/4)y = 4 (where (1/2), (1/4) and 4 are dollar amounts)

Let's eliminate y. To accomplish this, mult. the 2nd eq'n by-4, obtaining

-2x - y = -16, and then combining this with the 1st equation, as follows:

-2x - y = -16

x + y = 10

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

-x = -6, or x = 6. This represents the number of pens purchased by Terrance.

Subbing 6 for x in the first equation, we get y = 4. This represents the number of pencils he bought.

($0.50)x + ($0.25)y = $4.00 (dollar amounts)

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3 years ago

What does f(3)= ? <br> Please help answer me

Zielflug [23.3K]

Answer:

Step-by-step explanation:

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3 years ago

2 points) Sometimes a change of variable can be used to convert a differential equation y′=f(t,y) into a separable equation. One

Stells [14]

$y'=(t+y)^2-1$

Substitute $u=t+y$ , so that $u'=y'$ , and

$u'=u^2-1$

which is separable as

$\dfrac{u'}{u^2-1}=1$

Integrate both sides with respect to $t$ . For the integral on the left, first split into partial fractions:

$\dfrac{u'}2\left(\frac1{u-1}-\frac1{u+1}\right)=1$

$\displaystyle\int\frac{u'}2\left(\frac1{u-1}-\frac1{u+1}\right)\,\mathrm dt=\int\mathrm dt$

$\dfrac12(\ln|u-1|-\ln|u+1|)=t+C$

Solve for $u$ :

$\dfrac12\ln\left|\dfrac{u-1}{u+1}\right|=t+C$

$\ln\left|1-\dfrac2{u+1}\right|=2t+C$

$1-\dfrac2{u+1}=e^{2t+C}=Ce^{2t}$

$\dfrac2{u+1}=1-Ce^{2t}$

$\dfrac{u+1}2=\dfrac1{1-Ce^{2t}}$

$u=\dfrac2{1-Ce^{2t}}-1$

Replace $u$ and solve for $y$ :

$t+y=\dfrac2{1-Ce^{2t}}-1$

$y=\dfrac2{1-Ce^{2t}}-1-t$

Now use the given initial condition to solve for $C$ :

$y(3)=4\implies4=\dfrac2{1-Ce^6}-1-3\implies C=\dfrac3{4e^6}$

so that the particular solution is

$y=\dfrac2{1-\frac34e^{2t-6}}-1-t=\boxed{\dfrac8{4-3e^{2t-6}}-1-t}$

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3 years ago