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3 years ago

⚠️ATTENTION ⚠️ I need help with this please explain the problem -10x+9=-21

Mathematics

2 answers:

frutty [35]3 years ago

7 0

Answer:

x=3

Step-by-step explanation:

-10x+9=-21

Subtract 9 from each side

-10x+9-9=-21-9

-10x = -30

Divide each side by -10

-10x/-10 = -30/-10

x = 3

gizmo_the_mogwai [7]3 years ago

5 0

-10x + 9 = -21
-10x = -21 - 9
-10x = -30
x = -30 / -10
x = 3
Hope this help! If it is, Pls Brainliest

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For the given table of values for a polynomial function, where must the zeros of the function lie?

Jet001 [13]

Answer:

A. Between 3.0 and 3.5 and between 4.0 and 4.5

Step-by-step explanation:

The zeroes of a function occur whenever a value of x returns zero. To predict where the zeroes lie, determine the interval(s) where the function crosses the x-axis. This occurs when either $f(x)$ goes from a negative value to a positive value or vice versa.

From $x=3.0$ and $x=3.5$ , the y-values go from 4.0 (positive) to -0.2 (negative), respectively. Therefore, there must be a zero in this interval.

From $x=4.0$ and $x=4.5$ , the y-values go from -0.8 (negative) to 0.1 (positive), respectively. Therefore, there must also be a zero in this interval.

Thus, the zeros of this function occur between 3.0 and 3.5 and between 4.0 and 4.5, leading to answer choice A.

7 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:

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