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tatiyna
3 years ago
10

Ally purchased 215 pieces of candy for

Mathematics
1 answer:
Xelga [282]3 years ago
6 0

Answer: here is your answer she gave 208 friends her candy

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A shopping bag contains six orange juice boxes and four grape juice boxes. Without
juin [17]

Answer:

6/25 i think

Step-by-step explanation:

5 0
4 years ago
What is the slope of all lines parallel to the line 4x-5y+-1?
Bond [772]

Given equation of the line ,

\implies 4x - 5y +(-1) = 0

Simplify ,

\implies 4x - 5y - 1 = 0

Transpose terms to RHS ,

\implies 5y = 4x - 1

Divide both sides by 5 ,

\implies \boxed{ y =\dfrac{4}{5}x -\dfrac{1}{5}}

Now we know the slope intercept form of the line as y = mx + c we have ,

  • m =\dfrac{4}{5}

As we know that the slope of all parallel lines are same . Henceforth ,

  • m_{|| \ lines } =\dfrac{4}{5}
5 0
3 years ago
9/20
NISA [10]

For this case we must find the solution of the following quadratic equation:

x ^ 2 + 5x + 7 = 0

Where:

a = 1\\b = 5\\c = 7

Then, the solution is given by:

x = \frac {-b \pm \sqrt {b ^ 2-4 (a) (c)}} {2a}

Substituting the values:

x = \frac {-5 \pm \sqrt {5 ^ 2-4 (1) (7)}} {2 (1)}\\x = \frac {-5 \pm \sqrt {25-28}} {2}\\x = \frac {-5 \pm \sqrt {-3}} {2}

By definition we have:i ^ 2 = -1

x = \frac {-5 \pm \sqrt {3i ^ 2}} {2}\\x = \frac {-5 \pm i \sqrt {3}} {2}

Thus, we have two complex roots.

Answer:

The equation has no real roots.

4 0
4 years ago
Find a point-slope form for the line with slope 1/5<br> and passing through the point (-7,-4).
Basile [38]

Answer:

y + 4 = ⅕(x + 7)  

Step-by-step explanation:

The "point-slope" form of the equation of a straight line is:

y − y₁ = m(x − x₁)

where m is the slope and (x₁, y₁) is a point on the line.

If m = ⅕ and (x₁, y₁) = (-7,  - 4), the point-slope equation is

y + 4 = ⅕(x + 7)

The Figure below shows the graph of your line and a point at (-7, -4).

6 0
3 years ago
if f(x)=2(x)^2+5sqrt(x+2), complete the follwoing statement ( round your answer to the nearest hundreth) : f(0)=_____
Maslowich

Answer:

f(0)=7.07

Step-by-step explanation:

We have the function f(x)=2(x)^2+5\sqrt{(x+2)}

In this case we want to find the value of f0)

To find f(0) you must replace the x in the function with the number 0 and solve as shown below

f(0)=2(0)^2+5\sqrt{(0+2)}

f(0)=0+5\sqrt{(0+2)}

f(0)=5\sqrt{(2)}

Therefore

f(0)=7.07

5 0
4 years ago
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