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3 years ago

Tell me were the numbers go

Mathematics

1 answer:

Harrizon [31]3 years ago

7 0

A would go in rational numbers
B would go in integer
C would go in while numbers
D in irrational numbers
Hope this helps

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What is the length and the width of a rectangle with perimeter of 48 inches if its length is 7 inches longer than its width.

sattari [20]

Answer:

Length and the width of the rectangle is 15.5 inches and 8.5 inches respectively.

Step-by-step explanation:

Given:

Length = 7 inches + Width

=> P = 48 inches

=> L = (7 + W) inches............. (1)

Perimeter 'P' is the sum of all sides of the rectangle 2 (L + W)

∴ 48 inches = 2 (L + W) .................(2)

Substitute for L in equation (2)

∴ 48 inches = 2 [(7 + W) + W]

48 = 2[7 + 2W]

48 = 14 +4W

48 - 14 = 4W

34 = 4W

W = 8.5 inches

Recall that: Length, L = (7 + W)

∴ L = (7 + 8.5)

L = 15.5 inches

4 0

3 years ago

Read 2 more answers

A varies directly with B. If A=44 when B=110, find A when B=140

Kay [80]

$\frac{44}{a} = \frac{110}{140}$

$a = \frac{44 \times 140}{110} = 56$

7 0

3 years ago

At what x value does the function given below have a hole?<br><br> f(x)=x+3/x2−9

S_A_V [24]

Answer:

hole at x=-3

Step-by-step explanation:

The hole is the discontinuity that exists after the fraction reduces. (Still doesn't exist for original of course)

The discontinuities for this expression is when the bottom is 0. x^2-9=0 when x=3 or x=-3 since squaring either and then subtracting 9 would lead to 0.

So anyways we have (x+3)/(x^2-9)

= (x+3)/((x-3)(x+3))

Now this equals 1/(x-3) with a hole at x=-3 since the x+3 factor was "cancelled" from the denominator.

4 0

2 years ago

What is the median of the following numbers: 9, 10, 4, 6, 3, 7, 8, 2, 5. 54 8 7 6

qwelly [4]

The median of the numbers would be 7.5

8 0

3 years ago

Show work please<br> \sqrt(x+12)-\sqrt(2x+1)=1

Nesterboy [21]

Answer:

$x=4$

Step-by-step explanation:

Given $\displaystyle\\\sqrt{x+12}-\sqrt{2x+1}=1$ , start by squaring both sides to work towards isolating $x$ :

$\displaystyle\\\left(\sqrt{x+12}-\sqrt{2x+1}\right)^2=\left(1\right)^2$

Recall $(a-b)^2=a^2-2ab+b^2$ and $\sqrt{a}\cdot \sqrt{b}=\sqrt{a\cdot b}$ :

$\displaystyle\\\left(\sqrt{x+12}-\sqrt{2x+1}\right)^2=\left(1\right)^2\\\implies x+12-2\sqrt{(x+12)(2x+1)}+2x+1=1$

Isolate the radical:

$\displaystyle\\x+12-2\sqrt{(x+12)(2x+1)}+2x+1=1\\\implies -2\sqrt{(x+12)(2x+1)}=-3x-12\\\implies \sqrt{(x+12)(2x+1)}=\frac{-3x-12}{-2}$

Square both sides:

$\displaystyle\\(x+12)(2x+1)=\left(\frac{-3x-12}{-2}\right)^2$

Expand using FOIL and $(a+b)^2=a^2+2ab+b^2$ :

$\displaystyle\\2x^2+25x+12=\frac{9}{4}x^2+18x+36$

Move everything to one side to get a quadratic:

$\displaystyle-\frac{1}{4}x^2+7x-24=0$

Solving using the quadratic formula:

A quadratic in $ax^2+bx+c$ has real solutions $\displaystyle x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$ . In $\displaystyle-\frac{1}{4}x^2+7x-24$ , assign values:

$\displaystyle \\a=-\frac{1}{4}\\b=7\\c=-24$

Solving yields:

$\displaystyle\\x=\frac{-7\pm \sqrt{7^2-4\left(-\frac{1}{4}\right)\left(-24\right)}}{2\left(-\frac{1}{4}\right)}\\\\x=\frac{-7\pm \sqrt{25}}{-\frac{1}{2}}\\\\\begin{cases}x=\frac{-7+5}{-0.5}=\frac{-2}{-0.5}=\boxed{4}\\x=\frac{-7-5}{-0.5}=\frac{-12}{-0.5}=24 \:(\text{Extraneous})\end{cases}$

Only $x=4$ works when plugged in the original equation. Therefore, $x=24$ is extraneous and the only solution is $\boxed{x=4}$

4 0

2 years ago