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3 years ago

Is this correct??? I need help I’m not sure if it’s right...

Mathematics

1 answer:

Korolek [52]3 years ago

7 0

Answer:

Step-by-step explanation:

check it on a calculator to make sure it is correct because it look right to me.

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Matilda needs at least 112$ to buy a new dress . She has already saved $40 . She earns $9 an hour babysitting .write and solve a

WITCHER [35]

(112-40)/9=8 is the answer-er-er-er

7 0

3 years ago

Read 2 more answers

Please help me with the following questions. Thanks in advance!!

puteri [66]

Answer:

Step-by-step explanation:

In choices a and b, the bases are positive numbers greater than 1, and so these are growth functions. In c and d, the bases are between 0 and 1, and thus these are decay functions.

In the second problem we have 3ln(x + 1). Rewrite this as ln(x + 1)^3.

We also have 9ln(x - 4). Rewrite this as ln(x - 4)^9.

Because of the + sign connecting ln(x + 1)^3 and ln(x - 4)^9, these two logs combine to form

ln [ (x + 1)^3 ] * (x - 4)^9 (the log of a product).

Now we have:

ln [ (x + 1)^3 ] * (x - 4)^9 - 4ln(x + 7), or:

[ (x + 1)^3 ] * (x - 4)^9

ln ------------------------------------

(x + 7)^9

7 0

3 years ago

Read 2 more answers

Que is on pic.i can't able to type in text.

ad-work [718]

It's not difficult to compute the values of $A$ and $B$ directly:

$A=\displaystyle\int_1^{\sin\theta}\frac{\mathrm dt}{1+t^2}=\tan^{-1}t\bigg|_{t=1}^{t=\sin\theta}$
$A=\tan^{-1}(\sin\theta)-\dfrac\pi4$

$B=\displaystyle\int_1^{\csc\theta}\frac{\mathrm dt}{t(1+t^2)}=\int_1^{\csc\theta}\left(\frac1t-\frac t{1+t^2}\right)\,\mathrm dt$
$B=\left(\ln|t|-\dfrac12\ln|1+t^2|\right)\bigg|_{t=1}^{t=\csc\theta}$
$B=\ln\left|\dfrac{\csc\theta}{\sqrt{1+\csc^2\theta}}\right|+\dfrac12\ln2$

Let's assume $0$ , so that $|\csc\theta|=\csc\theta$ .

Now,

$\Delta=\begin{vmatrix}A&A^2&B\\e^{A+B}&B^2&-1\\1&A^2+B^2&-1\end{vmatrix}$
$\Delta=A\begin{vmatrix}B^2&-1\\A^2+B^2&-1\end{vmatrix}-e^{A+B}\begin{vmatrix}A^2&B\\A^2+B^2&-1\end{vmatrix}+\begin{vmatrix}A^2&B\\B^2&-1\end{vmatrix}$
$\Delta=A(-B^2+A^2+B^2)-e^{A+B}(-A^2-A^2B-B^3)+(-A^2-B^3)$
$\Delta=A^3-A^2-B^3+e^{A+B}(A^2+A^2B+B^3)$

There doesn't seem to be anything interesting about this result... But all that's left to do is plug in $A$ and $B$ .

3 0

2 years ago

Simplify fully please

neonofarm [45]

Answer:

8x² + 8

Step-by-step explanation:

Given

(x² + 3)² - (x² - 1)² ← expand both factors using FOIL

= $x^{4}$ + 6x² + 9 - ( $x^{4}$ - 2x² + 1) ← distribute by - 1

= $x^{4}$ + 6x² + 9 - $x^{4}$ + 2x² - 1 ← collect like terms

= 8x² + 8

8 0

3 years ago

Read 2 more answers

I need a little extra help with number three

diamong [38]

Answer:

The answer to your question is:

Step-by-step explanation:

Data

f(x) = -2x² + 8x - 2

Process

-2x² + 8x = y + 2

-2(x² - 4x + 4) = y + 2 - 8

-2(x - 2)² = y - 6

(x - 2)² = 1/2 (y - 6)

Vertex = (2, 6)

Axis of symmetry = x = 2

y-intercept

f(0) = -2(0)² + 8(0) - 2

f(0) = 0 + 0 - 2

f(0) = -2

Domain (-∞, ∞)

Range (-∞, 6]

See the graph below

6 0

2 years ago