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2 years ago

Is my answer correct? (Correct me if im wrong)

Mathematics

1 answer:

Tresset [83]2 years ago

8 0

Answer:

Step-by-step explanation:

I think u are right

I hope this helps

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Car X weighs 136 pounds more than car Z. Car Y weighs 117 pounds more than car Z. The total weight of all three cars is 9439 pou

Aleksandr [31]

Let x, y and z denote the weighs of car X, car Y and car Z, respectively.

We know that car X weighs 136 more than car Z, this can be express by the equation:

$x=z+136$

We also know that Y weighs 117 pounds more than car Z, this can be express as:

$y=z+117$

Finally, we know that the total weight of all the cars is 9439, then we have:

$x+y+z=9439$

Hence, we have the system of the equations:

$\begin{gathered} x=z+136 \\ y=z+117 \\ z+y+z=9439 \end{gathered}$

To solve the system we can plug the values of x and y, given in the first two equations, in the last equation; then we have:

$\begin{gathered} z+136+z+117+z=9439 \\ 3z=9439-136-117 \\ 3z=9186 \\ z=\frac{9186}{3} \\ z=3062 \end{gathered}$

Now that we have the value of z we plug it in the first two equations to find x and y:

$\begin{gathered} x=3062+136=3198 \\ y=3062+117=3179 \end{gathered}$

Therefore, car X weighs 3198 pound, car Y weighs 3179 pounds and car Z weighs 3062 pounds.

4 0

9 months ago

the happy play chargers $30 per hour for parties. Make a table of ordered pairs in which the x coordinate represents hours and t

Juli2301 [7.4K]

X:2,3,4,5.
Y:60,90,120,150

8 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

10) The ratio of boys to girls in Mrs. Ronilo's math class is 3 to 1. What PERCENT of the class is girls

klio [65]

Answer:

25%

Step-by-step explanation:

ratio is 3B : 1G (this means, out of 4 sections, 3 are boys and 1 are girls)

this means that 3/4 of class is boys and 1/4 are girls

1/4 equals 25%

7 0

1 year ago

Evaluate 5m2 for m=4

jolli1 [7]

If the question is 5m^2 the answer is 80. And if the question is 5m2 then it is 40

4 0

2 years ago

Read 2 more answers