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4 years ago

Alien shoestring is 0.73 meters long. Karen shoestring is 0.66 meter long. How much longer is alien shoestring than karen

Mathematics

1 answer:

Alja [10]4 years ago

3 0

0.07 meters longer .

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Which of the following statements are true of solving equations?

katrin2010 [14]

we will check each options

option-A:

For solving any equations , we always isolate variables on anyone side

For exp: x+7=1

so, this is TRUE

option-B:

For solving system of equations

For exp:

x-y=1

x+y=3

If we use addition , we could easily solve for x and y

so, this is TRUE

option-C:

We always solve problems using conventional method

we do not guess

so, this is FALSE

option-D:

We often reverse order of operation

For exp:

(x-2)^2-3=0

so, this is TRUE

option-E:

For linear equations , we always get one solution , infinite solutions or no solutions

so, this is FALSE

5 0

3 years ago

Read 2 more answers

Help with #17 please

ohaa [14]

2 evens is an positive
1 odd,1 even is a odd
2 odd is an ofd

8 0

3 years ago

The area of the triangle formed by x− and y− intercepts of the parabola y=0.5(x−3)(x+k) is equal to 1.5 square units. Find all p

Juliette [100K]

Check the picture below.

based on the equation, if we set y = 0, we'd end up with 0 = 0.5(x-3)(x-k).

and that will give us two x-intercepts, at x = 3 and x = k.

since the triangle is made by the x-intercepts and y-intercepts, then the parabola most likely has another x-intercept on the negative side of the x-axis, as you see in the picture, so chances are "k" is a negative value.

now, notice the picture, those intercepts make a triangle with a base = 3 + k, and height = y, where "y" is on the negative side.

let's find the y-intercept by setting x = 0 now,

$\bf y=0.5(x-3)(x+k)\implies y=\cfrac{1}{2}(x-3)(x+k)\implies \stackrel{\textit{setting x = 0}}{y=\cfrac{1}{2}(0-3)(0+k)} \\\\\\ y=\cfrac{1}{2}(-3)(k)\implies \boxed{y=-\cfrac{3k}{2}} \\\\[-0.35em] ~\dotfill\\\\ \stackrel{\textit{area of a triangle}}{A=\cfrac{1}{2}bh}~~ \begin{cases} b=3+k\\ h=y\\ \quad -\frac{3k}{2}\\ A=1.5\\ \qquad \frac{3}{2} \end{cases}\implies \cfrac{3}{2}=\cfrac{1}{2}(3+k)\left(-\cfrac{3k}{2} \right)$

$\bf \cfrac{3}{2}=\cfrac{3+k}{2}\left( -\cfrac{3k}{2} \right)\implies \stackrel{\textit{multiplying by }\stackrel{LCD}{2}}{3=\cfrac{(3+k)(-3k)}{2}}\implies 6=-9k-3k^2 \\\\\\ 6=-3(3k+k^2)\implies \cfrac{6}{-3}=3k+k^2\implies -2=3k+k^2 \\\\\\ 0=k^2+3k+2\implies 0=(k+2)(k+1)\implies k= \begin{cases} -2\\ -1 \end{cases}$

now, we can plug those values on A = (1/2)bh,

$\bf \stackrel{\textit{using k = -2}}{A=\cfrac{1}{2}(3+k)\left(-\cfrac{3k}{2} \right)}\implies A=\cfrac{1}{2}(3-2)\left(-\cfrac{3(-2)}{2} \right)\implies A=\cfrac{1}{2}(1)(3) \\\\\\ A=\cfrac{3}{2}\implies A=1.5 \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ \stackrel{\textit{using k = -1}}{A=\cfrac{1}{2}(3+k)\left(-\cfrac{3k}{2} \right)}\implies A=\cfrac{1}{2}(3-1)\left(-\cfrac{3(-1)}{2} \right) \\\\\\ A=\cfrac{1}{2}(2)\left( \cfrac{3}{2} \right)\implies A=\cfrac{3}{2}\implies A=1.5$

7 0

3 years ago

How can I create a word problem with the following expressions 15+2d 200-4m

Alenkasestr [34]

For the first expression
15 + 2d
A possible word problem would be this:
A person saves $2 per day of money. Before he started saving, he had $15 dollars set aside. Look for the expression that expresses the total amount of money saved in terms of the number of days passed

The second expression is
200 - 2m
A possible word problem would be this:
The distance from school to the park is 200m. A kid riding a bike is traveling at a speed of 2m/s from the school to the park. Write an expression for the distance remaining between the park and the kid.

6 0

3 years ago

Please help need answers

bogdanovich [222]

see explanation below

(1) $\frac{1}{5}$ × $\frac{2}{2}$ = $\frac{2}{10}$ = 0.2