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babymother [125]
3 years ago
8

What is the answer for 5 more than a number y is -2

Mathematics
2 answers:
Rudik [331]3 years ago
4 0
Y + 5 = -2               Mathematical equation that represents this problem.

y = -7                     Subtract 5 from each side.

Answer: y = -7
faltersainse [42]3 years ago
4 0
Try using 'let statements' to start this problem of.

let y = 5 more than -2

now, let's make an equation.

5+y= -2

then,just solve for y by subtracting 5 from both sides of the equation.

y=-7
as a final step, you just need to check.
5+y= -2

5+-7= -2
-2=-2 ✔


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Sue had 5 identical bags of candy to share with the crew. Twenty-five pieces of candy disappeared. Ted had 4 bags with the same
inna [77]

Answer:

Each bag originally had 35 pieces of candy

Step-by-step explanation:

Since the bags are identical, Let them Number of candies in each bag sue had be x. Thus, she had 5x candies. Since she lost 25 pieces, then she now has;

5x - 25 candies left.

We are told that Ted had 4 bags with the same number of candies like Sue had. Thus, he has 4x candies.

He mysteriously found 10 additional pieces. Total Ted now has is;. 4x + 10

Finally, we are told that Sue and Ted now have an equal number of candies.

Thus;

5x - 25 = 4x + 10

Rearranging, we have;

5x - 4x = 25 + 10

x = 35

Thus, each bag originally had 35 pieces of candy

3 0
3 years ago
Inequalities? the product of 11 and a number is less than 53 i'm confused on how to write it out
nordsb [41]
Ok, so we know that the product of means something added, and we are given that eleven and a number are less than 53. So the equation is 11+X<53.
3 0
3 years ago
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erastovalidia [21]
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3 0
3 years ago
Steven is driving to Texas pulling a 12 ft. boat. If he drives 150 miles in
ziro4ka [17]

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Step-by-step explanation:

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6 0
3 years ago
Find the two intersection points
bogdanovich [222]

Answer:

Our two intersection points are:

\displaystyle (3, -2) \text{ and } \left(-\frac{53}{25}, \frac{46}{25}\right)

Step-by-step explanation:

We want to find where the two graphs given by the equations:

\displaystyle (x+1)^2+(y+2)^2 = 16\text{ and } 3x+4y=1

Intersect.

When they intersect, their <em>x-</em> and <em>y-</em>values are equivalent. So, we can solve one equation for <em>y</em> and substitute it into the other and solve for <em>x</em>.

Since the linear equation is easier to solve, solve it for <em>y: </em>

<em />\displaystyle y = -\frac{3}{4} x + \frac{1}{4}<em />

<em />

Substitute this into the first equation:

\displaystyle (x+1)^2 + \left(\left(-\frac{3}{4}x + \frac{1}{4}\right) +2\right)^2 = 16

Simplify:

\displaystyle (x+1)^2 + \left(-\frac{3}{4} x  + \frac{9}{4}\right)^2 = 16

Square. We can use the perfect square trinomial pattern:

\displaystyle \underbrace{(x^2 + 2x+1)}_{(a+b)^2=a^2+2ab+b^2} + \underbrace{\left(\frac{9}{16}x^2-\frac{27}{8}x+\frac{81}{16}\right)}_{(a+b)^2=a^2+2ab+b^2} = 16

Multiply both sides by 16:

(16x^2+32x+16)+(9x^2-54x+81) = 256

Combine like terms:

25x^2+-22x+97=256

Isolate the equation:

\displaystyle 25x^2 - 22x -159=0

We can use the quadratic formula:

\displaystyle x = \frac{-b\pm\sqrt{b^2-4ac}}{2a}

In this case, <em>a</em> = 25, <em>b</em> = -22, and <em>c</em> = -159. Substitute:

\displaystyle x = \frac{-(-22)\pm\sqrt{(-22)^2-4(25)(-159)}}{2(25)}

Evaluate:

\displaystyle \begin{aligned} x &= \frac{22\pm\sqrt{16384}}{50} \\ \\ &= \frac{22\pm 128}{50}\\ \\ &=\frac{11\pm 64}{25}\end{aligned}

Hence, our two solutions are:

\displaystyle x_1 = \frac{11+64}{25} = 3\text{ and } x_2 = \frac{11-64}{25} =-\frac{53}{25}

We have our two <em>x-</em>coordinates.

To find the <em>y-</em>coordinates, we can simply substitute it into the linear equation and evaluate. Thus:

\displaystyle y_1 = -\frac{3}{4}(3)+\frac{1}{4} = -2

And:

\displaystyle y _2 = -\frac{3}{4}\left(-\frac{53}{25}\right) +\frac{1}{4} = \frac{46}{25}

Thus, our two intersection points are:

\displaystyle (3, -2) \text{ and } \left(-\frac{53}{25}, \frac{46}{25}\right)

6 0
3 years ago
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