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3 years ago

Please hurry I need an answer for this quiz

Mathematics

2 answers:

Temka [501]3 years ago

8 0

Answer:

Step-by-step explanation:

Lapatulllka [165]3 years ago

3 0

Answer:

I think the answer would be C.

Step-by-step explanation:

hope I help

You might be interested in

Which graph shows the solution to the following inequality? 4d- 6 < 18 PLEASE HELP

kenny6666 [7]

Answer:

answer a

Step-by-step explanation:

$4d-6$ < $18$

Let's isolate $d$ on one side of the equation. Ignore the inequality for now. We'll deal with that later.

$4d-6=18\\4d-6+6=18+6\\4d=24\\\frac{4d}{4}=\frac{24}{4}\\ d=6$

Now, I'm going to bring back the inequality or < symbol. I only removed it when simplifying and isolating $d$ , but if this confuses you, just do your math and keep the inequality there.

$d$ < $6$

On a number line, this would include every number less than $6$ , due to the < (less than) symbol. This disqualifies answers b and d because they are showing every number greater than $6$ . But, how do we decide between answers a and c?
If a line has point at its beginning, $6$ , then that means that every number less than or equal to [ ≤ ] 6 is being shown, but our equation just says less than [ < ] 6, so answer a is our correct answer.

4 0

3 years ago

Read 2 more answers

50 Benson purchased a $1,800 certificate of deposit that earns s

Pavlova-9 [17]

Answer:

3.2%

Step-by-step explanation:

The computation of the annual rate of interest is shown below:

As we know that

Simple interest = Principal × rate of interest × time period

($1,857.60 - $1,800) = $1,800 × rate of interest × 1

$57.60 = $1,800 × rate of interest × 1

So, the rate of interest is

= $57.60 ÷ $1,800 × 100

= 3.2%

Hence, the rate of interest on annual basis is 3.2%

5 0

2 years ago

PLEASE HELP I DO NOT UNDERSTAND AT ALL ITS PRECALC PLEASE SERIOUS ANSWERS

Elina [12.6K]

You want to end up with $A\sin(\omega t+\phi)$ . Expand this using the angle sum identity for sine:

$A\sin(\omega t+\phi)=A\sin(\omega t)\cos\phi+A\cos(\omega t)\sin\phi$

We want this to line up with $2\sin(4\pi t)+5\cos(4\pi t)$ . Right away, we know $\omega=4\pi$ .

We also need to have

$\begin{cases}A\cos\phi=2\\A\sin\phi=5\end{cases}$

Recall that $\sin^2x+\cos^2x=1$ for all $x$ ; this means

$(A\cos\phi)^2+(A\sin\phi)^2=2^2+5^2\implies A^2=29\implies A=\sqrt{29}$

Then

$\begin{cases}\cos\phi=\frac2{\sqrt{29}}\\\sin\phi=\frac5{\sqrt{29}}\end{cases}\implies\tan\phi=\dfrac{\sin\phi}{\cos\phi}=\dfrac52\implies\phi=\tan^{-1}\left(\dfrac52\right)$

So we end up with

$2\sin(4\pi t)+5\cos(4\pi t)=\sqrt{29}\sin\left(4\pi t+\tan^{-1}\left(\dfrac52\right)\right)$

8 0

3 years ago

Read 2 more answers

1+-w2+9w and I need help cuz I’m on 76 and I’m sooo close help

Gnesinka [82]

$\huge \boxed{\mathfrak{Question} \downarrow}$

Simplify :- 1 + - w² + 9w.

$\large \boxed{\mathfrak{Answer \: with \: Explanation} \downarrow}$

$\large \sf1 + - w ^ { 2 } + 9 w$

Quadratic polynomial can be factored using the transformation $\sf \: ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right)$ , where $\sf x_{1} and x_{2}$ are the solutions of the quadratic equation $\sf \: ax^{2}+bx+c=0$ .

$\large \sf-w^{2}+9w+1=0$

All equations of the form $\sf\:ax^{2}+bx+c=0$ can be solved using the quadratic formula: $\sf\frac{-b±\sqrt{b^{2}-4ac}}{2a}$ . The quadratic formula gives two solutions, one when ± is addition and one when it is subtraction.

$\large \sf \: w=\frac{-9±\sqrt{9^{2}-4\left(-1\right)}}{2\left(-1\right)} \\$

Square 9.

$\large \sf \: w=\frac{-9±\sqrt{81-4\left(-1\right)}}{2\left(-1\right)} \\$

Multiply -4 times -1.

$\large \sf \: w=\frac{-9±\sqrt{81+4}}{2\left(-1\right)} \\$

Add 81 to 4.

$\large \sf \: w=\frac{-9±\sqrt{85}}{2\left(-1\right)} \\$

Multiply 2 times -1.

$\large \sf \: w=\frac{-9±\sqrt{85}}{-2} \\$

Now solve the equation $\sf\:w=\frac{-9±\sqrt{85}}{-2}$ when ± is plus. Add -9 to $\sf\sqrt{85}$ .

$\large \sf \: w=\frac{\sqrt{85}-9}{-2} \\$

Divide -9+ $\sf\sqrt{85}$ by -2.

$\large \boxed{ \sf \: w=\frac{9-\sqrt{85}}{2}} \\$

Now solve the equation $\sf\:w=\frac{-9±\sqrt{85}}{-2}$ when ± is minus. Subtract $\sf\sqrt{85}$ from -9.

$\large \sf \: w=\frac{-\sqrt{85}-9}{-2} \\$

Divide $\sf-9-\sqrt{85}$ by -2.

$\large \boxed{ \sf \: w=\frac{\sqrt{85}+9}{2}} \\$

Factor the original expression using $\sf\:ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right)$ . Substitute $\sf\frac{9-\sqrt{85}}{2}$ for $\sf\:x_{1}$ and $\sf\frac{9+\sqrt{85}}{2}$ for $\sf\:x_{2}$ .

$\large \boxed{ \boxed {\mathfrak{-w^{2}+9w+1=-\left(w-\frac{9-\sqrt{85}}{2}\right)\left(w-\frac{\sqrt{85}+9}{2}\right) }}}$

Well, in the picture you inserted it says that it's 8th grade mathematics. So, I'm not sure if you have learned simplification with the help of biquadratic formula. So, if you want the answer simplified only according to like terms then your answer will be ⇨

$\large \sf \: 1 + - w {}^{2} + 9w \\ =\large \boxed{\bf \: 1 - {w}^{2} + 9w}$

This cannot be further simplified as there are no more like terms (you can use the biquadratic formula if you've learned it.)

4 0

2 years ago

What is the volume of the square pyramid with bad edges 12cm and height 12cm?

spayn [35]

The volume of a pyramid is found by multiplying the length, width and height together then dividing by 3.

length x width x height = 12 x 12 x 12 = 1728

volume = 1728 / 3 = 576 cubic cm

8 0

3 years ago