How to explain 9/100+7/10=

Mathematics

2 answers:

Sladkaya [172]3 years ago

7 0

The answer is 0.709
9/100 = 0.09
7/10 = 0.7
So this answer is 0.79

Advocard [28]3 years ago

5 0

First you need to find a common denominator.

in this case it would be 100.

so now you have 9/100 + 70/100

then you add the numerators and keep the denominator the same

9/100 + 70/100 = 79/100

you can not simplify this so that is your final answer

You might be interested in

Point P is located at (−2, 7), and point R is located at (1, 0). Find the y value for the point Q that is located two over three

otez555 [7]

Answer:

$y_Q=\dfrac{21}{5}=4.2$

Step-by-step explanation:

If the point Q that is located two over three the distance from point P to point R, then PQ:QR=2:3.

Use formula to find the coordinates of the point Q:

$x_Q=\dfrac{3x_P+2x_R}{3+2}\\ \\y_Q=\dfrac{3y_P+2y_R}{3+2}$

In your case, P(-2,7) and R(1,0), then

$x_Q=\dfrac{3\cdot (-2)+2\cdot 1}{3+2}=\dfrac{-4}{5}\\ \\y_Q=\dfrac{3\cdot 7+2\cdot 0}{3+2}=\dfrac{21}{5}$

6 0

3 years ago

Which lines are perpendicular to the line y - 1 = 1/3 (x+2)? Check all that apply.

ioda

Answer: Options 1, 3, 5

Step-by-step explanation:

Perpendicular lines have slopes that are negative reciprocals, so since the slope of the given line is 1/3, we need to find lines with a slope of -3.

Option 1 has a slope of -3.
Option 2 has a slope of 3.
Option 3 has a slope of -3.
Option 4 has a slope of 1/3.
If we subtract 3x from both sides, we get y=-3x+7, so option 5 has a slope of -3.

7 0

2 years ago

Suppose a parabola has an axis of symmetry at x=-5, a maximum height of 9, and passes through the point (-7,1). Write the equati

Nutka1998 [239]

the parabola has maximum at 9, meaning is a vertical parabola and it opens downwards.

it has a symmetry at x = -5, namely its vertex's x-coordinate is -5.

check the picture below.

so then, we can pretty much tell its vertex is at (-5 , 9), and we also know it passes through (-7, 1)

$\bf ~~~~~~\textit{parabola vertex form} \\\\ \begin{array}{llll} y=a(x- h)^2+ k\qquad \leftarrow \textit{using this one}\\\\ x=a(y- k)^2+ h \end{array} \qquad\qquad vertex~~(\stackrel{}{ h},\stackrel{}{ k}) \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ \begin{cases} h=-5\\ k=9 \end{cases}\implies y=a[x-(-5)]^2+9\implies y=a(x+5)^2+9$

$\bf \textit{we also know that } \begin{cases} x=-7\\ y=1 \end{cases}\implies 1=a(-7+5)^2+9 \\\\\\ -8=a(-2)^2\implies -8=4a\implies \cfrac{-8}{4}=a\implies -2=a \\\\[-0.35em] ~\dotfill\\\\ ~\hfill y=-2(x+5)^2+9~\hfill$