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quester [9]
3 years ago
8

A jar of jam contains 55% fruit.How much fruit would I need for ten 454g jars of jam?​

Mathematics
2 answers:
Svetach [21]3 years ago
5 0
10x454x0.55=2,497 grams of fruit
dolphi86 [110]3 years ago
3 0

Answer: 10x454x0.55=2,497 grams of fruit

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Find a vector parametrization for the line with the given description. Perpendicular to the yz-plane, passes through (0, 0, 8)
Brilliant_brown [7]

Answer:

\left\{\begin{array}{l}x=t\\ \\y=0\\ \\z=8\end{array}\right.

or

\overrightarrow{(t,0,8)}

Step-by-step explanation:

yz-plane has the equation x=a, where a is a real constant. The normal vector of this plane (vector perpendicular to the plane) is

\overrightarrow {n}=(1,0,0)

If the line is perpendicular to this plane, then it is parallel to the normal vector, so the equation of this line is

\dfrac{x-x_0}{1}=\dfrac{y-y_0}{0}=\dfrac{z-z_0}{0},

where (x_0,y_0,z_0) are coordinates of the point the line is passing through.

In your case, the line is passing through the point (0,0,8), so the canonical equation of the line is

\dfrac{x-0}{1}=\dfrac{y-0}{0}=\dfrac{z-8}{0}

Write a vector parametrization for this line

\left\{\begin{array}{l}\dfrac{x-0}{1}=t\\ \\\dfrac{y-0}{0}=t\\ \\\dfrac{z-8}{0}=t\end{array}\right.\Rightarrow \left\{\begin{array}{l}x=t\\ \\y=0\\ \\z=8\end{array}\right.

3 0
3 years ago
Factor<br> 2z to the second power +3z-9
kozerog [31]
The answer is (2z-3)(z+3)
5 0
3 years ago
Please help giving 10pts
MArishka [77]

Answer:

x=0   x=9

Step-by-step explanation:

f(x) = x^3 - 9x^2

Set equal to zero to find the zeros

0= x^3 - 9x^2

Factor out x^2

0=x^2 (x-9)

Using the zero product property

0=x^2   0=x-9

x=0   x=9

8 0
3 years ago
Read 2 more answers
Write the polar equation in rectangular form.
zhuklara [117]

Answer:

Step-by-step explanation:

The identities you need here are:

r=\sqrt{x^2+y^2}  and   r^2=x^2+y^2

You also need to know that

x = rcosθ  and

y = rsinθ

to get this done.

We have

r = 6 sin θ

Let's first multiply both sides by r (you'll always begin these this way; you'll see why in a second):

r² = 6r sin θ

Now let's replace r² with what it's equal to:

x² + y² = 6r sin θ

Now let's replace r sin θ with what it's equal to:

x² + y² = 6y

That looks like the beginnings of a circle.  Let's get everything on one side because I have a feeling we will be completing the square on this:

x^2+y^2-6y=0

Complete the square on the y-terms by taking half its linear term, squaring it and adding it to both sides.

The y linear term is 6.  Half of 6 is 3, and 3 squared is 9, so we add 9 in on both sides:

x^2+(y^2-6y+9)=9

In the process of completing the square, we created within that set of parenthesis a perfect square binomial:

x^2+(y-3)^2=9

And there's your circle!  Third choice down is the one you want.

Fun, huh?

5 0
2 years ago
Suppose theta is an angle in the standard position whose terminal side is in quadrant 4 and cot theta = -6/7. find the exact val
zimovet [89]

First off, let's notice that the angle is in the IV Quadrant, where sine is negative and the cosine is positive, likewise the opposite and adjacent angles respectively.

Also let's bear in mind that the hypotenuse is never negative, since it's simply just a radius unit.

\bf cot(\theta )=\cfrac{\stackrel{adjacent}{6}}{\stackrel{opposite}{-7}}\qquad \impliedby \textit{let's find the \underline{hypotenuse}} \\\\\\ \textit{using the pythagorean theorem} \\\\ c^2=a^2+b^2\implies c=\sqrt{a^2+b^2} \qquad \begin{cases} c=hypotenuse\\ a=adjacent\\ b=opposite\\ \end{cases} \\\\\\ c=\sqrt{6^2+(-7)^2}\implies c=\sqrt{36+49}\implies c=\sqrt{85} \\\\[-0.35em] ~\dotfill

\bf tan(\theta)=\cfrac{\stackrel{opposite}{-7}}{\stackrel{adjacent}{6}} ~\hfill csc(\theta)=\cfrac{\stackrel{hypotenuse}{\sqrt{85}}}{\stackrel{opposite}{-7}} ~\hfill sec(\theta)=\cfrac{\stackrel{hypotenuse}{\sqrt{85}}}{\stackrel{adjacent}{6}} \\\\\\ sin(\theta)=\cfrac{\stackrel{opposite}{-7}}{\stackrel{hypotenuse}{\sqrt{85}}}\implies \stackrel{\textit{and rationalizing the denominator}}{sin(\theta)=\cfrac{-7}{\sqrt{85}}\cdot \cfrac{\sqrt{85}}{\sqrt{85}}\implies sin(\theta)=-\cfrac{7\sqrt{85}}{85}}

\bf cos(\theta)=\cfrac{\stackrel{adjacent}{6}}{\stackrel{hypotenuse}{\sqrt{85}}}\implies \stackrel{\textit{and rationalizing the denominator}}{cos(\theta)=\cfrac{6}{\sqrt{85}}\cdot \cfrac{\sqrt{85}}{\sqrt{85}}\implies cos(\theta)=\cfrac{6\sqrt{85}}{85}}

6 0
2 years ago
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