All categories

4 years ago

Thirteen greater than a number is 192. Find the number. Check your solution.

Mathematics

2 answers:

VARVARA [1.3K]4 years ago

8 0

Answer:

179

Step-by-step explanation:

vlabodo [156]4 years ago

5 0

Answer:

Thirteen greater than 192- the number is 179.

Solution:

Let the number be x. It is given that the thirteen greater than a number is 192. Since we have assumed the number to be x, thus we have to find the value of x.

Using the below equation we will try to find the value of x.

$13+x = 192$

$x = 192 -13$

$x = 179$

Thus the number is 179.

To check the solution we put the value of x in the left hand side (L.H.S) equation and match it with the right hand side value.

$L.H.S = 13+x$

$\Rightarrow 13+179$

$\Rightarrow 192 = Right-hand side$

You might be interested in

What do you think you need to do to the

8090 [49]

Answer:

You need to multiply the terms together, therefore adding the exponents 2 and 3 to get 5.

Step-by-step explanation:

5 0

3 years ago

Can i get help on this question x+4= -9 1/4

Bas_tet [7]

Answer:
x=-25/4
Step-by-step explanation:

First your make both sides clean,

x+4=-9/4

Then you minus 4 on both sides to balance the equation:

x+4-4=(-9/4)-4
x= -25/4

6 0

3 years ago

Evaluate the line integral, where c is the given curve. (x + 9y) dx + x2 dy, c c consists of line segments from (0, 0) to (9, 1)

viktelen [127]

$\displaystyle\int_C(x+9y)\,\mathrm dx+x^2\,\mathrm dy=\int_C\langle x+9y,x^2\rangle\cdot\underbrace{\langle\mathrm dx,\mathrm dy\rangle}_{\mathrm d\mathbf r}$

The first line segment can be parameterized by $\mathbf r_1(t)=\langle0,0\rangle(1-t)+\langle9,1\rangle t=\langle9t,t\rangle$ with $0\le t\le1$ . Denote this first segment by $C_1$ . Then

$\displaystyle\int_{C_1}\langle x+9y,x^2\rangle\cdot\mathbf dr_1=\int_{t=0}^{t=1}\langle9t+9t,81t^2\rangle\cdot\langle9,1\rangle\,\mathrm dt$
$=\displaystyle\int_0^1(162t+81t^2)\,\mathrm dt$
$=108$

The second line segment ( $C_2$ ) can be described by $\mathbf r_2(t)=\langle9,1\rangle(1-t)+\langle10,0\rangle t=\langle9+t,1-t\rangle$ , again with $0\le t\le1$ . Then

$\displaystyle\int_{C_2}\langle x+9y,x^2\rangle\cdot\mathrm d\mathbf r_2=\int_{t=0}^{t=1}\langle9+t+9-9t,(9+t)^2\rangle\cdot\langle1,-1\rangle\,\mathrm dt$
$=\displaystyle\int_0^1(18-8t-(9+t)^2)\,\mathrm dt$
$=-\dfrac{229}3$

Finally,

$\displaystyle\int_C(x+9y)\,\mathrm dx+x^2\,\mathrm dy=108-\dfrac{229}3=\dfrac{95}3$

5 0

3 years ago

I need step by step instructions on how to get the y-intercept for the following function:

Svetllana [295]

Luktdkilfdiykfg;oizxkuilouyk7it8rl,kvkuc kugc ulv lug

4 0

4 years ago

The amount of water dispensed by a water dispenser is normally distributed,

Contact [7]

Answer:

The correct answer is - C. 11.45 ounces to 11.75 ounces.

Step-by-step explanation:

According to the empirical rule of the distribution for 68% falls under the normal curve falls within 1 standard deviation of the mean.

That is:

μ±δ

From the given information, the mean is

μ = 11.60

and the standard deviation is

δ = 0.15

We substitute the given parameters to obtain;

11.60±0.15

11.75 and 11.45

This means the lower limit is

11.45

and the upper limit is

11.75

3 0

3 years ago