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3 years ago

Needing Help!!! Please Answer Fast! :)

Mathematics

2 answers:

zimovet [89]3 years ago

8 0

Pls help me I stuck and I’m only a 4 grader

NeTakaya3 years ago

6 0

Answer:

Step-by-step explanation:

Use the vertical line text. If any vertical line intersect the relation at more than one point, it is not a function. So, using this method, you get C.

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0=4t-16t^2 Solve this please

vaieri [72.5K]

Answer:

$\large\boxed{t=0\ \vee\ t=\dfrac{1}{4}}$

Step-by-step explanation:

$4t-16t^2=0\qquad\text{divide both sides by 4}\\\\\dfrac{4t}{4}-\dfrac{16t^2}{4}=\dfrac{0}{4}\\\\t-4t^2=0\\\\t(1-4t)=0\iff t=0\ \vee\ 1-4t=0\\\\1-4t=0\qquad\text{subtract 1 from both sides}\\\\-4t=-1\qquad\text{divide both sides by (-4)}\\\\t=\dfrac{1}{4}$

7 0

3 years ago

Please someone answer this, “what is the area of triangle ABC?”

prisoha [69]

Here's the solution,

The given triangle is an equilateral triangle, because it's all angles measure 60° each

now,

Area of an equilateral triangle :

=》

$\dfrac{ \sqrt{3} }{4} {a}^{2}$

where,

a = 8 ( side measure )

=》

$\dfrac{ \sqrt{3} }{4} \times 8 \times 8$

=》

$\sqrt{3} \times 16$

=》

$16 \sqrt{3} \: \: \: cm {}^{2}$

=》

$27.71 \: \: cm {}^{2}$

4 0

2 years ago

For number 6, evaluate the definite integral.

maks197457 [2]

$\bf \displaystyle \int\limits_{0}^{28}\ \cfrac{1}{\sqrt[3]{(8+2x)^2}}\cdot dx\impliedby \textit{now, let's do some substitution}\\\\
-------------------------------\\\\
u=8+2x\implies \cfrac{du}{dx}=2\implies \cfrac{du}{2}=dx\\\\
-------------------------------\\\\$

$\bf \displaystyle \int\limits_{0}^{28}\ \cfrac{1}{\sqrt[3]{u^2}}\cdot \cfrac{du}{2}\implies \cfrac{1}{2}\int\limits_{0}^{28}\ u^{-\frac{2}{3}}\cdot du\impliedby 
\begin{array}{llll}
\textit{now let's change the bounds}\\
\textit{by using } u(x)
\end{array}\\\\
-------------------------------\\\\
u(0)=8+2(0)\implies u(0)=8
\\\\\\
u(28)=8+2(28)\implies u(28)=64$

$\bf \\\\
-------------------------------\\\\
\displaystyle \cfrac{1}{2}\int\limits_{8}^{64}\ u^{-\frac{2}{3}}\cdot du\implies \cfrac{1}{2}\cdot \cfrac{u^{\frac{1}{3}}}{\frac{1}{3}}\implies \left. \cfrac{3\sqrt[3]{u}}{2} \right]_8^{64}
\\\\\\
\left[ \cfrac{3\sqrt[3]{(2^2)^3}}{2} \right]-\left[ \cfrac{3\sqrt[3]{2^3}}{2} \right]\implies \cfrac{12}{2}-\cfrac{6}{2}\implies 6-3\implies 3$

3 0

3 years ago

The three sides of a triangle are: 16, 12 and 22. Is the triangle Acute, Right, or Obtuse?

tamaranim1 [39]

I know it’s not right, I think it’s obtuse

7 0

3 years ago

If f(x) = x2 + x, find f(-3). -12 3 6

ANEK [815]

Answer:

-9

Step-by-step explanation:

-3(2) + (-3)

-6 + (-3)

-9

4 0

2 years ago

Read 2 more answers