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2 years ago

Someone please help me

Mathematics

2 answers:

juin [17]2 years ago

8 0

Answer:

1/1 = 1

Step-by-step explanation:

Use rise/run.

The x value increases by 1 in the same amount of time that the y value increases by 1. This gives you the slope 1/1 (y increase/x increase).

1/1 can then be converted to 1.

goldenfox [79]2 years ago

5 0

Answer:

the slope is -2

Step-by-step explanation:

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joe weighs 20lbs.less yhantwice jeffs weight.if jeff would gain 10lbs.,then together they wouldweigh 250lbs. how much does each

masya89 [10]

So,

Let x represent Joe's weight and y represent Jeff's weight.

"Joe weighs 20 lbs. less than twice Jeff's weight."
x = 2y - 20

"If Jeff would gain 10 lbs., then together they would weigh 250 lbs."
(y + 10) + x = 250

We now have our two open sentences.
x = 2y - 20
(y + 10) + x = 250

Get rid of parentheses.
x = 2y - 20
x + y + 10 = 250

We will use Elimination by Substitution.
2y - 20 + y + 10 = 250

Collect Like Terms.
3y - 10 = 250

Add 10 to both sides.
3y = 260

Divide both sides by 3.
$Jeff's\ weight = 86 \frac{2}{3} \ lbs.$

Substitute again.
$x = 2(86 \frac{2}{3} )-20$

Multiply.
$x = 173 \frac{1}{3} -20$

Subtract.
$Joe's\ weight = 153 \frac{1}{3}\ lbs.$

Check.

"Joe weighs 20 lbs. less than twice Jeff's weight."
Jeff's weight times two is 173 and one-third.
20 lbs. less than that is 153 and one-third lbs. Check.

"If Jeff would gain 10 lbs., then together they would weigh 250 lbs."
86 and two-thirds + 10 = 96 and two-thirds.
96 and two-thirds + 153 and one-third equals 250 lbs. Check.

$Joe's\ weight\ is\ 153 \frac{1}{3} \ lbs.$
$Jeff's\ weight\ is\ 86 \frac{2}{3} \ lbs.$

3 0

3 years ago

Amy had homework for math,science,and history.she spent 1/3 of her time working on math and 1/4 of her time working on science.

Alja [10]

First we need to multiply to make the fractions have equal denominators so we can add/subtract.

math: 1/3 x 4/4=4/12
science: 1/4 x 3/3=3/12

Now we can find the time she spent on history (I assume this is what you are looking for)

Total time-math time-science time=history time
12/12-4/12-3/12
=5/12

Amy spent 5/12 of her time working on history.

8 0

3 years ago

Read 2 more answers

PLEASE HELP!!! Find the equation , in the standard form of the line passing through the points (3,-4) and (5,1)

ExtremeBDS [4]

$\bf \begin{array}{ccccccccc}
&&x_1&&y_1&&x_2&&y_2\\
% (a,b)
&&(~ 3 &,& -4~) 
% (c,d)
&&(~ 5 &,& 1~)
\end{array}
\\\\\\
% slope = m
slope = m\implies 
\cfrac{\stackrel{rise}{ y_2- y_1}}{\stackrel{run}{ x_2- x_1}}\implies \cfrac{1-(-4)}{5-3}\implies \cfrac{1+4}{5-3}\implies \cfrac{5}{2}
\\\\\\
% point-slope intercept
\stackrel{\textit{point-slope form}}{y- y_1= m(x- x_1)}\implies y-(-4)=\cfrac{5}{2}(x-3)\implies y+4=\cfrac{5}{2}x-\cfrac{15}{2}$

$\bf y=\cfrac{5}{2}x-\cfrac{15}{2}-4\implies y=\cfrac{5}{2}x-\cfrac{23}{2}\impliedby 
\begin{array}{llll}
\textit{now let's multiply both}\\
\textit{sides by }\stackrel{LCD}{2}
\end{array}
\\\\\\
2(y)=2\left( \cfrac{5}{2}x-\cfrac{23}{2} \right)\implies 2y=5x-23\implies \stackrel{standard~form}{-5x+2y=-23}
\\\\\\
\textit{and if we multiply both sides by -1}\qquad 5x-2y=23$

side note: multiplying by the LCD of both sides is just to get rid of the denominators

5 0

3 years ago

H(x)=x^2-4x+1 find axis of symmetry

Tresset [83]

Answer:

the answer is x=2

I used math-way

7 0

3 years ago

Zaid has a peculiar pair of four-sided dice. When he rolls the dice, the probability of any particular outcome is proportional t

Elden [556K]

Answer: a) $\bold{\dfrac{3}{16}}$ b) $\bold{\dfrac{1}{36}}$

Step-by-step explanation:

a) In order to get an even number, you have the 3 different scenarios:

1) Even, Even, Even, Even $\dfrac{3\times 3\times 3\times 3}{6^4} \quad = \dfrac{3^4}{6^4}\quad =\dfrac{1}{16}$

2) Even, Even, Odd, Odd $\dfrac{3\times 3\times 3\times 3}{6^4} \quad = \dfrac{3^4}{6^4}\quad =\dfrac{1}{16}$

3) Odd, Odd, Odd, Odd $\dfrac{3\times 3\times 3\times 3}{6^4} \quad = \dfrac{3^4}{6^4}\quad =\dfrac{1}{16}$

Order doesn't matter

Add them up to get your answer: $\dfrac{1}{16}+\dfrac{1}{16}+\dfrac{1}{16}\quad =\large\boxed{\dfrac{3}{16}}$

b) If one die is a 2 and another is a 3 and the other two dice can be any number, then you have 1 possibility for a 2, 1 possibility for a 3, and 6 possibilities for each of the other two dice.

$\dfrac{1\times 1\times 6\times 6}{6^4}\quad =\dfrac{1}{6^2}\quad =\large\boxed{\dfrac{1}{36}}$

3 0

3 years ago