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3 years ago

PleaseI really need help on this please

Mathematics

2 answers:

Karo-lina-s [1.5K]3 years ago

4 0

Answer:

Step-by-step explanation:

mamaluj [8]3 years ago

3 0

Answer:

Step-by-step explanation:

the anwser to number 7 is d or (x2-4) (x2+4)

by the way try using the app called Photo Math i use it for my test it works greatyou should try it some day.

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Lily used a coordinate plane to graph the path she walked from her house to the library. She started at her house at point A(–6,

natka813 [3]

Answer:The answer is 19

Step-by-step explanation:

Count the units. WH

en you count all of them they give you 19 :D

6 0

2 years ago

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If Kathy walks 3 miles in 80 minutes how far will she in 120 minutes

zepelin [54]

Answer:

4.5 miles

Step-by-step explanation:

We can use a ratio to solve

3 miles x miles

--------------- = ----------------

80 minutes 120 minutes

Using cross products

3 * 120 = 80x

Divide each side by 80

360/80 = 80x/80

4.5 =x

8 0

3 years ago

Which is the completely factored form of 18x4 – 42x3 + 24x2?

Serga [27]

Answer:

Step-by-step explanation:

5 0

3 years ago

The complement of an angle is 6 degrees less than twice the measure of the angle , . Find the angle and its supplement .

katrin [286]

9514 1404 393

Answer:

angle: 32°
supplement: 148°

Step-by-step explanation:

Let x represent the angle. Then its complement is 2x-6:

x +(2x -6) = 90

3x = 96

x = 32

180 -x = 148

The angle is 32°; its supplement is 148°.

_____

Additional comment

We are given a relation between the angle and its complement. We are asked for the value of the angle and its supplement. A typo in the problem is suspected.

5 0

2 years ago

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