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3 years ago

Solving this: 7x +7 * 1 =?

Mathematics

1 answer:

Alecsey [184]3 years ago

6 0

The answer in simplest form is: 7x +7

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During the football game, Justin caught 3 passes. One was for a touchdown and was a 52 yard gain. The other was for a first down

pogonyaev

Answer:

Step-by-step explanation:

52+17-10=59

5 0

3 years ago

Find the value of each variable.

Harrizon [31]

Step-by-step explanation:

b is per the identity of angles on parallel lines when intersected by one inclined line the same as the 40° angle.

so,

b = 40°

due to the parallel nature of the 2 lines there is a symmetry effect for such shapes inscribed a circle. the upper and the lower triangle must be similar. and when applying a vertical line through the central crossing point, everything to the left is mirrored by everything on the right.

so, angle c must be equal to angle b.

c = 40°

and as the sum of all angles in a triangle is always 180°, d is then

d = 180 - 40 - 40 = 100°

the interior angle of the arc angle a is the supplementary angle of d (together they are 180°), because together with d they cover the full down side of the top-left to bottom-right line.

interior angle to a = 180 - 100 = 80°

due to the symmetry again, the arc angle opposite to a is the same as a.

as we know, the interior angle to a pair of opposing arc angles is the mean value of the 2 angles.

so, we have

(a + a)/2 = 80

2a/2 = 80

a = 80°

there might (and actually should) be some more direct approaches for "a" out of the other pieces of information, but that was the most straight one right out of my mind, and I don't spend time on finding additional shortcuts, when I have already a working approach.

8 0

1 year ago

A spherical balloon is inflated so that its volume is increasing at the rate of 3 ft3/min. How fast is the diameter of the ballo

beks73 [17]

When an spherical balloon volume is increasing at the rate of $3ft^3/min$ then the diameter of the balloon is increasing $\frac{3}{2\pi }ft /min$

How can we find the rate of change of balloon's diameter ?

The volume of a spherical balloon is $v=\frac{4}{3} \pi r^3$

In form of diameter we can write as

$v=\frac{4}{3} \pi (\frac{D}{2} )^3\\=\frac{1}{6} \pi D^3$

Now we will differentiate both sides wrt to $t$ we get

$\frac{dv}{dt} =\frac{1}{6} \pi 3D^2 \frac{dD}{dt} \\\frac{dD}{dt} =\frac{2}{\pi D^2} \frac{dv}{dt} \\\\when r=1\\D=2ft$

Given in the question $\frac{dv}{dt} =3ft^3/min$

thus when we substitute the values we get

$\frac{dD}{dt} =\frac{2}{\pi *2^2} (3)\\\frac{dD}{dt}=\frac{3}{2\pi } ft/min$

Learn more about the differentiation here:

brainly.com/question/28046488

#SPJ4

3 0

2 years ago

Using radicals, what is an equivalent expression for the expression 2 1/3

faltersainse [42]

Answer:

Step-by-step explanation:

21/3

3 will council it self one and 21 7times

4 0

3 years ago

interpret r(t) as the position of a moving object at time t. Find the curvature of the path and determine thetangential and norm

Igoryamba

Answer:

The curvature is $\kappa=1$

The tangential component of acceleration is $a_{\boldsymbol{T}}=0$

The normal component of acceleration is $a_{\boldsymbol{N}}=1 (2)^2=4$

Step-by-step explanation:

To find the curvature of the path we are going to use this formula:

$\kappa=\frac{||d\boldsymbol{T}/dt||}{ds/dt}$

where

$\boldsymbol{T}}$ is the unit tangent vector.

$\frac{ds}{dt}=|| \boldsymbol{r}'(t)}||$ is the speed of the object

We need to find $\boldsymbol{r}'(t)$ , we know that $\boldsymbol{r}(t)=cos \:2t \:\boldsymbol{i}+sin \:2t \:\boldsymbol{j}+ \:\boldsymbol{k}$ so

$\boldsymbol{r}'(t)=\frac{d}{dt}\left(cos\left(2t\right)\right)\:\boldsymbol{i}+\frac{d}{dt}\left(sin\left(2t\right)\right)\:\boldsymbol{j}+\frac{d}{dt}\left(1)\right\:\boldsymbol{k}\\\boldsymbol{r}'(t)=-2\sin \left(2t\right)\boldsymbol{i}+2\cos \left(2t\right)\boldsymbol{j}$

Next , we find the magnitude of derivative of the position vector

$|| \boldsymbol{r}'(t)}||=\sqrt{(-2\sin \left(2t\right))^2+(2\cos \left(2t\right))^2} \\|| \boldsymbol{r}'(t)}||=\sqrt{2^2\sin ^2\left(2t\right)+2^2\cos ^2\left(2t\right)}\\|| \boldsymbol{r}'(t)}||=\sqrt{4\left(\sin ^2\left(2t\right)+\cos ^2\left(2t\right)\right)}\\|| \boldsymbol{r}'(t)}||=\sqrt{4}\sqrt{\sin ^2\left(2t\right)+\cos ^2\left(2t\right)}\\\\\mathrm{Use\:the\:following\:identity}:\quad \cos ^2\left(x\right)+\sin ^2\left(x\right)=1\\\\|| \boldsymbol{r}'(t)}||=2\sqrt{1}=2$

The unit tangent vector is defined by

$\boldsymbol{T}}=\frac{\boldsymbol{r}'(t)}{||\boldsymbol{r}'(t)||}$

$\boldsymbol{T}}=\frac{-2\sin \left(2t\right)\boldsymbol{i}+2\cos \left(2t\right)\boldsymbol{j}}{2} =\sin \left(2t\right)+\cos \left(2t\right)$

We need to find the derivative of unit tangent vector

$\boldsymbol{T}'=\frac{d}{dt}(\sin \left(2t\right)\boldsymbol{i}+\cos \left(2t\right)\boldsymbol{j}) \\\boldsymbol{T}'=-2\cdot(\sin \left(2t\right)\boldsymbol{i}+\cos \left(2t\right)\boldsymbol{j})$