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3 years ago

Uhhh I am struggling with this question algebra 1

Mathematics

1 answer:

Olin [163]3 years ago

4 0

Answer:

-2x^2 +16x -8

Step-by-step explanation:

Oh s-

Like terms i guess Same variable you can substract. Remember you only substract from 1 number not from all.

-2x^2 stays the same as there is no like term

9x - (-7x)

-2 - (6)

-2x^2 + 9x+ 7x -2-6

-2x^2 +16x -8

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Please answer will see about brainliest

mina [271]

it is d. Pensacola beach

3 0

3 years ago

Consider the vectors a =3i +j −k, b =i +j +4k, c=i +3j +k, d =−i −3j +k, g =−3i −j +k. Which pairs (if any) of these vectors are

11111nata11111 [884]

Answer:

a and b are perpendicular to each other, as are b and d, b and g

Step-by-step explanation:

To check whether two vectors are perpendicular to each other, we need the angle between these vectors to be 90 degrees.

We can find the angle between to vectors a and b from the following relation:

The cosine of the angle $\theta$ between two vectors is equal to the dot product of this vectors divided by the product of vector magnitude.

$cos(\theta) = \frac{a.b}{|a||b|}$

cos(90) = 0, so when the dot product between vectors a and b is 0, it means that these vectors are perpendicular to each other.

Now, for your exercise, let's compute the dot product between these vectors.

-----------

a.b = (3,1,-1).(1,1,4) = 3+1-4 = 0

So a and b are perpendicular to each other.

------------

a.c = (3,1,-1).(1,3,1) = 3+3-1 = 5

a and c are not perpendicular to each other.

--------------

a.d = (3,1,-1).(-1,-3,1) = -3-3-1 = -7

So not perpendicular

---------------

a.g = (3,1,-1).(-3,-1,1) = -9-1-1 = -11

Not

-----------------

b.c = (1,1,4).(1,3,1) = 1+3+4 = 8

Not

------------------

b.d = (1,1,4).(-1,-3,1) = -1 -3 +4 = 0

b.d = 0, so b and d are perpendicular to each other

--------------------

b.g = (1,1,4).(-3,-1,1) = -3-1+4 = 0

b.g = 0, perpendicular

---------------------

c.d = (1,3,1).(-1,-3,1) = -1-9+1 = -9

-----------------------

c.g = (1,3,1).(-3,-1,1) = -3-3+1 = -5

------------------------

d.g = (-1,-3,1).(-3,-1,1) = 3+3+1 = 7

7 0

3 years ago

Dilate the given points distance from the origin (x,y) (0.8x, 0.8y) . The point (-10,-20) becomes & point (15,25) become

dalvyx [7]

Answer: The point (-10,-20) becomes (-8,-16) & point (15,25) becomes (12,20).

Step-by-step explanation:

When a point (x,y) is dilated by a scale factor k from the origin, then the new points become

$(kx,ky).$

The given rule of dilation : (x,y) →(0.8x, 0.8y)

The first point is (-10,-20), so its image will be

$(-10,-20)\to (0.8\times-10,0.8\times-20)=(-8,-16)$

So, the point (-10,-20) becomes (-8,-16).

The second point is (15,25), so its image will be

$(15,25)\to (0.8\times15,0.8\times25)=(12,20)$

So, the point (15,25) becomes (12,20).

Hence, the point (-10,-20) becomes (-8,-16) & point (15,25) becomes (12,20).

8 0

3 years ago

Find an equation for G.

zaharov [31]

Answer:

i think the answer is 60

Step-by-step explanation:

5 0

3 years ago

Use the given information to find the exact value of the trigonometric function

eimsori [14]

$\begin{gathered} \csc \theta=-\frac{6}{5} \\ \tan \theta>0 \\ \cos \frac{\theta}{2}=\text{?} \end{gathered}$

Half Angle Formula

$\cos \frac{\theta}{2}=\pm\sqrt[\square]{\frac{1+\cos\theta}{2}}$ $\tan \theta>0\text{ and csc}\theta\text{ is negative in the third quadrant}$ $\begin{gathered} \csc \theta=-\frac{6}{5}=\frac{r}{y} \\ x^2+y^2=r^2 \\ x=\pm\sqrt[\square]{r^2-y^2} \\ x=\pm\sqrt[\square]{6^2-(-5)^2} \\ x=\pm\sqrt[\square]{36-25} \\ x=\pm\sqrt[\square]{11} \\ \text{x is negative since the angle is on the 3rd quadrant} \end{gathered}$ $\begin{gathered} \cos \theta=\frac{x}{r}=\frac{-\sqrt[\square]{11}}{6} \\ \cos \frac{\theta}{2}=\pm\sqrt[\square]{\frac{1+\cos\theta}{2}} \\ \cos \frac{\theta}{2}is\text{ also negative in the 3rd quadrant} \\ \cos \frac{\theta}{2}=-\sqrt[\square]{\frac{1+\frac{-\sqrt[\square]{11}}{6}}{2}} \\ \cos \frac{\theta}{2}=-\sqrt[\square]{\frac{\frac{6-\sqrt[\square]{11}}{6}}{2}} \\ \cos \frac{\theta}{2}=-\sqrt[\square]{\frac{6-\sqrt[\square]{11}}{12}} \\ \\ \end{gathered}$

Answer:

$\cos \frac{\theta}{2}=-\sqrt[\square]{\frac{6-\sqrt[\square]{11}}{12}}$

Checking:

$\begin{gathered} \frac{\theta}{2}=\cos ^{-1}(-\sqrt[\square]{\frac{6-\sqrt[\square]{11}}{12}}) \\ \frac{\theta}{2}=118.22^{\circ} \\ \theta=236.44^{\circ}\text{ (3rd quadrant)} \end{gathered}$

Also,

$\csc \theta=\frac{1}{\sin\theta}=\frac{1}{\sin (236.44)}=-\frac{6}{5}\text{ QED}$

The answer is none of the choices

7 0

1 year ago