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lbvjy [14]
3 years ago
13

For a fundraiser, a group plans to sell granola bars and bottles of water at the same prices as described in Part A. The group w

ants the income from the fundraiser to be at least $150.
Choose the inequality to show the number of granola bars x and the number of bottles of water y that must be sold.

A. 1.4x + 0.75 y > 150
B. 1.4x + 0.75y ≥ 150
C. 1.4x + 0.75 < 150
D. 1.4x + 0.75 ≤ 150
Mathematics
2 answers:
shutvik [7]3 years ago
6 0

Answer:

100

Step-by-step explanation:

goblinko [34]3 years ago
4 0

Answer:

A

Step-by-step explanation:

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795 rounded to the nearest million
Andre45 [30]
I’m guessing it is 1 million because you can only round up since it’s value is not above 1 million.
3 0
3 years ago
The number of goals scored by a soccer team varies directly as the number of shots on goal. If the Lakeside Bears scored 4 goals
enot [183]

Since it varies directly, divide the number of shots by the number of goals:

48 shots / 4 goals = 12

This tells you they score a goal every 12 shots.

Multiply shots needed by number of goals:

12 x 3 = 36

They need 36 more shots to score 3 more goals.

6 0
3 years ago
The graph shows y as a function of x
nirvana33 [79]
I'm guessing Q lol. I hope that's helpful.
3 0
3 years ago
Part B what are the probabilities of each outcome in the sample space? select all that apply
zepelin [54]

Answer:

C, D, and E are correct

Step-by-step explanation:

p(2)= 1/6; p(3)= 1/6; p(4)= 1/6; 1/6=1/6=1/6

p(1)= 3/6; 3/6=1/2

p(4) = 1/6; There are six sections and one section is labeled<em> '4' </em>

<em />

Hope this helped! ;p

7 0
3 years ago
Use the Chain Rule to find the indicated partial derivatives. z = x^4 + xy^3, x = uv^4 + w^3, y = u + ve^w Find : ∂z/∂u , ∂z/∂v
k0ka [10]

I'll use subscript notation for brevity, i.e. \frac{\partial f}{\partial x}=f_x.

By the chain rule,

z_u=z_xx_u+z_yy_u

z_v=z_xx_v+z_yy_v

z_w=z_xx_w+z_yy_w

We have

z=x^4+xy^3\implies\begin{cases}z_x=4x^3+y^3\\z_y=3xy^2\end{cases}

and

\begin{cases}x=uv^4+w^3\\y=u+ve^w\end{cases}\implies\begin{cases}x_u=v^4\\x_v=4uv^3\\x_w=3w^2\\y_u=1\\y_v=e^w\\y_w=ve^w\end{cases}

When u=1,v=1,w=0, we have

\begin{cases}x(1,1,0)=1\\y(1,1,0)=2\end{cases}\implies\begin{cases}z_x(1,2)=12\\z_y(1,2)=12\end{cases}

and the partial derivatives take on values of

\begin{cases}x_u(1,1,0)=1\\x_v(1,1,0)=4\\x_w(1,1,0)=0\\y_u(1,1,0)=1\\y_v(1,1,0)=1\\y_w(1,1,0)=1\end{cases}

So we end up with

\boxed{\begin{cases}z_u(1,1,0)=24\\z_v(1,1,0)=60\\z_w(1,1,0)=12\end{cases}}

3 0
3 years ago
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