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inna [77]
2 years ago
12

Whoever finishes this gets "brainliest".just for fun35a^7--11a^7

Mathematics
1 answer:
8090 [49]2 years ago
4 0

Answer:

a=0

Step-by-step explanation:

Factor:

a^7 (35-11)

Set it equal to zero and solve:

a^7=0

a=0

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3 to the 4th power + 2 ⋅ 5 =
Mandarinka [93]
Hello,

The answer is "x=91".

Reason:

First write the equation:

3^4+2*5=

Now use PEMDAS:

E goes first:

3^4=3*3*3*3=81

81+2*5

M goes next:

2*5=10

81+10=

Now use A:

81+10=91

x=91

If you need anymore help feel free to ask me!

Hope this helps!

~Nonportrit
4 0
2 years ago
How to do this here
Andru [333]
Refer to the attached photo to see the problem worked out
8 0
3 years ago
Given that triangle x y z is congruent to triangle n m l, the measure of angle x equals 20 degrees, the measure of angle m equal
Delvig [45]

A triangle has a total angle of 180 degrees. Since we know that angle X = angle N = 20 degrees and angle Y = angle M = 75 degrees, therefore angle Z or angle L is equal to:

angle L = 180 – 20 – 75

<span>angle L =  85 degrees</span>

 

Therefore w is:

angle L = 5 (w – 2)

5 (w – 2) = 85

w – 2 = 17

<span>w = 19</span>

7 0
3 years ago
1.
creativ13 [48]
Let x represent the worth of the professional basketball player's autograph a year ago. Since his autograph's worth increased by 40% now, $364 is 1.40x. 
 
                                      364 = 1.40x

Solving for x gives x = 260.

Thus, the autograph of the basketball player was worth only $260 last year. 

4 0
2 years ago
I can't figure out how to do (i + j) x (i x j)for vector calc
Vinil7 [7]

In three dimensions, the cross product of two vectors is defined as shown below

\begin{gathered} \vec{A}=a_1\hat{i}+a_2\hat{j}+a_3\hat{k} \\ \vec{B}=b_1\hat{i}+b_2\hat{j}+b_3\hat{k} \\ \Rightarrow\vec{A}\times\vec{B}=\det (\begin{bmatrix}{\hat{i}} & {\hat{j}} & {\hat{k}} \\ {a_1} & {a_2} & {a_3} \\ {b_1} & {b_2} & {b_3}\end{bmatrix}) \end{gathered}

Then, solving the determinant

\Rightarrow\vec{A}\times\vec{B}=(a_2b_3-b_2a_3)\hat{i}+(b_1a_3+a_1b_3)\hat{j}+(a_1b_2-b_1a_2)\hat{k}

In our case,

\begin{gathered} (\hat{i}+\hat{j})=1\hat{i}+1\hat{j}+0\hat{k} \\ \text{and} \\ (\hat{i}\times\hat{j})=(1,0,0)\times(0,1,0)=(0)\hat{i}+(0)\hat{j}+(1-0)\hat{k}=\hat{k} \\ \Rightarrow(\hat{i}\times\hat{j})=\hat{k} \end{gathered}

Where we used the formula for AxB to calculate ixj.

Finally,

\begin{gathered} (\hat{i}+\hat{j})\times(\hat{i}\times\hat{j})=(1,1,0)\times(0,0,1) \\ =(1\cdot1-0\cdot0)\hat{i}+(0\cdot0-1\cdot1)\hat{j}+(1\cdot0-0\cdot1)\hat{k} \\ \Rightarrow(\hat{i}+\hat{j})\times(\hat{i}\times\hat{j})=1\hat{i}-1\hat{j} \\ \Rightarrow(\hat{i}+\hat{j})\times(\hat{i}\times\hat{j})=\hat{i}-\hat{j} \end{gathered}

Thus, (i+j)x(ixj)=i-j

8 0
1 year ago
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