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4 years ago

It’s IXL I absolutely suck at it, please help!!

Mathematics

2 answers:

Ket [755]4 years ago

7 0

Answer:

$\huge \boxed{s^{-28} }$

Step-by-step explanation:

$s^{-4} \cdot s^{-24}$

Applying exponent rule : $a^b \cdot a^c=a^{b+c}$

$s^{-4} \cdot s^{-24} = s^{-4+-24}=s^{-28}$

Rom4ik [11]4 years ago

3 0

S to the power of -28

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Will give BRAINLIEST to correct answer.<br><br> Given that ℓ1 ∥ ℓ2, find x.

lisov135 [29]

X = 38

2x is equal to the angle next to the x + 66 angle because ℓ1 ∥ ℓ2 is parallel so:

2x + x + 66 = 180 -- combine like terms

3x + 66 = 180 -- subtract 66 from both sides

3x = 114 -- divide by 3 on both side

x = 38

5 0

3 years ago

The graph of an exponential function passes through (2,−32) and (3,−64). Find the exponential function that describes the graph.

Greeley [361]

Answer: $y=(-8)2^x$

Step-by-step explanation:

The exponential function will have this form:

$y=ab^x$

We know that the function passes through the points $(2,-32)$ and $(3,-64)$ . Then, we can substitute the coordinates of the point $(2,-32)$ into $y=ab^x$ and solve for "a":

$-32=ab^2\\\\a=\frac{-32}{b^2}$

Then, we know that:

$y=(\frac{-32}{b^2})b^x$

Now, we neeed to substitute the coordinates of the second point $(3,-64)$ into $y=(\frac{-32}{b^2})b^x$ and solve for "b":

$-64=(\frac{-32}{b^2})b^3\\\\-64=-32b\\\\\frac{-64}{-32}=b\\\\b=2$

Substituting the value of "b" into $a=\frac{-32}{b^2}$ we can find "a":

$a\frac{-32}{2^2}\\\\a=-8$

Therefore, we get that the exponential function that describes the graph, is:

$y=(-8)2^x$

7 0

3 years ago

X=y+2

Andreas93 [3]

Answer:

4=8 908575939835

Step-by-step explanation:

lol

8 0

4 years ago

Find all the points having an x-coordinate of 4 whose distance from the point (-2,-1) is 10

Marta_Voda [28]

$A(x_1;\ y_1);\ B(x_2;\ y_2)\\\\the\ distance\ between\ A\ and\ B:\\\\d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}\\\\A(-2;-1);\ B(4;\ y);\ d=10\\\\subtitute\\\\\sqrt{(4-(-2))^2+(y-(-1))^2}=10\\\\\sqrt{(4+2)^2+(y+1)^2}=10\\\\\sqrt{6^2+(y+1)^2}=10\\\\\sqrt{36+(y+1)^2}=10\ \ \ |square\ both\ sides\\\\36+(y+1)^2=10^2\\\\36+(y+1)^2=100\ \ \ |subtract\ 36\ from\ both\ sides\\\\(y+1)^2=64\iff y+1=-\sqrt{64}\ or\ y+1=\sqrt{64}\\\\y+1=-8\ or\ y+1=8\ \ \ \ |subtract\ 1\ from\ both\ sides\\\\y=-9\ or\ y=7$

$Answer:{\boxed{(4;-9)\ or\ (4;\ 7)}$

4 0

3 years ago

choose the equation that represents a line that represents a line that passes through points -3, 2 and 2,1 A. 5x+y=-13 B. 5x-y=1

Mrrafil [7]

bearing in mind that standard form for a linear equation means

• all coefficients must be integers, no fractions

• only the constant on the right-hand-side

• all variables on the left-hand-side, sorted

• "x" must not have a negative coefficient

$\bf (\stackrel{x_1}{-3}~,~\stackrel{y_1}{2})\qquad (\stackrel{x_2}{2}~,~\stackrel{y_2}{1}) \\\\\\ slope = m\implies \cfrac{\stackrel{rise}{ y_2- y_1}}{\stackrel{run}{ x_2- x_1}}\implies \cfrac{1-2}{2-(-3)}\implies \cfrac{-1}{2+3}\implies -\cfrac{1}{5} \\\\\\ \begin{array}{|c|ll} \cline{1-1} \textit{point-slope form}\\ \cline{1-1} \\ y-y_1=m(x-x_1) \\\\ \cline{1-1} \end{array}\implies y-2=-\cfrac{1}{5}[x-(-3)]\implies y-2=-\cfrac{1}{5}(x+3)$

$\bf y-2=-\cfrac{1}{5}x-\cfrac{3}{5}\implies \stackrel{\textit{multiplying both sides by }\stackrel{LCD}{5}}{5\left(y-2 \right)=5\left( -\cfrac{1}{5}x-\cfrac{3}{5} \right)}\implies 5y-10=-x-3 \\\\\\ 5y=-x+7\implies x+5y=7$

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4 years ago