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Pepsi [2]
3 years ago
13

Scaling Question 2 Which describes the product when two fractions greater than 0 and less than 1 are multiplied? Your answer: Th

e product is less than both factors. The product is greater than one factor and less than the other factor. The product is greater than both factors, but less than 1. The product is greater than both factors and greater than 1. Clear answer edu o СМ​
Mathematics
1 answer:
qaws [65]3 years ago
3 0

The product is greater than  both factors but less than 1.

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Why can't the graphing calculator be used to solve (or approximate solutions to) all polynomial equations?
ipn [44]
Because it is normally set to only find real roots, not complex roots

if you are smart, you can use the factor function and use your head to find those roots
5 0
3 years ago
Pasta Place needs to make pasta with garlic cream sauce for 80 guests at a party. One batch of pasta calls for 3 cups of cream a
uysha [10]

Answer:

Total pints of cream will they need for the pasta = 7 pints

Step-by-step explanation:

Given - Pasta Place needs to make pasta with garlic cream sauce for 80 guests at a party. One batch of pasta calls for 3 cups of cream and feeds 10 people. The restaurant already has 5 pints of cream in the fridge.

To find - How many more pints of cream will they need for the pasta?

Solution -

10 people = 3 cups of cream

⇒80 people = 8×3 = 24 cups of cream

Now,

We know that,

1 cup = 0.5 pints

⇒24 cup = 24×0.5 = 12 pints.

Now,

Given that, The restaurant already has 5 pints of cream in the fridge.

So,

Total pints more needed = 12 - 5 = 7 pints.

∴ we get

Total pints of cream will they need for the pasta = 7 pints

3 0
3 years ago
32 is very much need to be answered
vova2212 [387]

Question 32):  

<u>Least To Greatest:</u>

-7/2,    -2.8,   -5/4,  4/3,  1.3

<u>Number of Inputs:   </u>

<u>5</u>

<u>Ascending Order (Least to Greatest)/Smallest to Largest:  </u>

-7/2,    -2.8,   -5/4,  1.3,  4/3

 

<u>Descending Order (Greatest to Least/Largest to Smallest):</u>

4/3, 1.3, -5/4, -2.8, -7/2





Hope that helps!!!                                         : )



3 0
3 years ago
Solve the following system. y = (1/2)x 2 + 2x - 1 and 3x - y = 1 The solutions are ( )and ( ) (remember to include the commas)
frozen [14]
\begin{cases}y=\dfrac{1}{2}x+2x-1\\3x-y=1\end{cases}\\\\\\&#10;\begin{cases}y=\dfrac{5}{2}x-1\3x-y=1\end{cases}\\\\\\&#10;3x-\left(\dfrac{5}{2}x-1\right)=1\\\\3x-\dfrac{5}{2}x+1=1\\\\&#10;3x-\dfrac{5}{2}x=0\\\\\dfrac{x}{2}=0\\\\x=0\\\\3\times0-y=1\\0-y=1\\y=-1\\\\\boxed{(x,y)=(0,-1)}

<span>The solutions are (0) and (-1)</span>
8 0
3 years ago
Read 2 more answers
It is known that the straight line l is tangent to the circle x^2+y^2=4 at a point on the x-axis, and intersects with the straig
Talja [164]

Step-by-step explanation:

The general equation of a circle is

                                         (x \ - \ h)^2 \ + \ (y \ - \ k)^2 \ = \ r^{2},

where <em>h</em> and <em>k</em> forms the coordinates of the centre of the circle.

When the circle has a centre at the origin, the equation reduces into

                        .                            x^2 \ + \ y^2 \ = r^2.

Now, we are interested in solving for the <em>x</em>-intercepts (the <em>x</em>-coordinates when the circle intersects the <em>x</em>-axis), of the circle

                                                      x^2 \ + \ y^2 \ = \ 4 .

Thus,

                                                     x^2 \ + \ (0)^2 \ = \ 4 \\ \\ \\ \-\hspace{1.3cm} x^{2} \ = \ 4 \\ \\ \\ \-\hspace{1.4cm} x \ = \ \pm \ 2.

Geometrically speaking, the tangent to the circle at the point defined by one of the <em>x</em>-intercepts of the circle is actually a vertical line, more specifically the lines x \ = \ \pm \ 2.

First and foremost, for the vertical line x \ = \ 2, it intersects the straight line y \ = \ \displaystyle\frac{1}{2}x  , giving the y-coordinate for point P,

                                                    y \ = \ \displaystyle\frac{1}{2}(2) \\ \\ \\ y \ = \ 1.

Hence, the coordinates of point P are (1, \ 1).

However, since there are no boundaries given in the question and a circle is symmetrical about its centre, thus, point P also exists when the vertical line x \ = \ -2 and interdects the straight line y \ = \ \displaystyle\frac{1}{2}x.

                                                      y \ = \ \displaystyle\frac{1}{2}(-2) \\ \\ \\ y \ = \ -1.

Therefore, the coordinates of point P are also (1, \ -1).

8 0
2 years ago
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