All categories

3 years ago

My answer is correct ?

Mathematics

1 answer:

vredina [299]3 years ago

8 0

Answer:

Yes

Step-by-step explanation:

You might be interested in

What is the first step in solving the equation x2 - 16 =0?

JulijaS [17]

Answer:

is that supposed to be 2x-16=0 cause if it is then

Step-by-step explanation:

2x-16=0

add 16

2x=16

divide by 2

x=16

5 0

3 years ago

What is the value of each 5 in 35252

Reil [10]

The first would be 5,000 and the second is 50

5 0

4 years ago

A concession stand sold hamburgers, salted pretzels, and hot dogs at a basketball game. Hamburgers were sold for $3.00, salted p

andreyandreev [35.5K]

Answer:

110

Step-by-step explanation:

Set hamburgers as h,

salted pretzels as p,

hot dogs as d.

h+p+d=144

(3.00)h+(2.25)p+(2.00)d=354.25

=>3h+2.25p+2d=354.25

we also know (2.00)d=82 => d=41

We make 3h+2.25p+2d=354.25 => 3h+2.25p=272.25

and h+p+d=144 => h+p=103

Set h=p-103

Substitution

3(p-103)+2.25p=272.25

3p-309+2.25p=272.25

(5.25)p=581.25

p=581.25/5.25=110.7142857

We round down =>110.

5 0

3 years ago

PLEASE HELP VERY URGENT** Write this verbal expression into an actual expression: The quotient of a number and 5 is 44.

OlgaM077 [116]

Answer: n/5=44

probably

Step-by-step explanation:

a quotient is the answer to a division problem and it asks for the answer to a division problem of a number(n) and 5 so it would have to be n/5 and then just equal to 44. A way to solve this would be to multiply 44 and 5 which is 220 which equals n

7 0

4 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

3 years ago