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erastova [34]
3 years ago
15

Help someone how do you solve this

Mathematics
1 answer:
Flauer [41]3 years ago
5 0

Answer:

a = 10/3

Step-by-step explanation:

This can be written as:

x^{2} (x^{4})^{1/3}  = x^{a}

Anything to the power of 1/3 means to find the cube root. We can simplify this to:

x^{2} (x^{4/3}) = x^{a}

I just multiplied 4 and 1/3 (because of the second index law (a^{m})^{n}  = a^{mn}). Now, we can add the powers because the bases are the same and we are multiplying (first index law):

x^{10/3} = x^{a}

a = 10/3

Hope this helps!

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The sum of two numbers is 15. Using x to represent the smaller of the two numbers, translate "the sum of twice the smaller numbe
Paladinen [302]

Answer:

Step-by-step explanation: Let x be the smaller and y be the largest number.

Since x+y=13, we deduce y=13-x

Now, for the translation: "two more than the larger number" is y+2 , while "twice the smaller" is 2x

Their sum is y+2+2x

And since we know that y=13-x, we have y+2+2x=13-x+2+2=15+x

4 0
1 year ago
The n term of a geometric sequence is denoted by Tn and the sum of the first n terms is denoted by Sn.Given T6-T4=5/2 and S5-S3=
Leno4ka [110]
1 step: S_{5}=T_{1}+T_{2}+T_{3}+T_{4}+T_{5}, S_{3}=T_{1}+T_{2}+T_{3}, then
 S_{5}-S_{3}=T_{4}+T_{5}=5.

2 step: T_{n}=T_{1}*q^{n-1}, then 
T_{6}=T_{1}*q^{5}
T_{5}=T_{1}*q^{4}
T_{4}=T_{1}*q^{3}
T_{3}=T_{1}*q^{2}
and \left \{ {{T_{6}-T_{4}= \frac{5}{2} } \atop {T_{5}+T_{4}=5}} \right. will have form \left \{ {{T_1*q^{5}-T_{1}*q^{3}= \frac{5}{2} } \atop {T_{1}*q^{4}+T_{1}*q^{3}=5} \right..

3 step: Solve this system  \left \{ {{T_1*q^{3}*(q^{2}-1)= \frac{5}{2} } \atop {T_{1}*q^{3}*(q+1)=5} \right. and dividing first equation on second we obtain \frac{q^{2}-1}{q+1}= \frac{ \frac{5}{2} }{5}. So, \frac{(q-1)(q+1)}{q+1} = \frac{1}{2} and q-1= \frac{1}{2}, q= \frac{3}{2} - the common ratio.

4 step: Insert q= \frac{3}{2}into equation T_{1}*q^{3}*(q+1)=5 and obtain T_{1}* \frac{27}{8}*( \frac{3}{2}+1 ) =5, from where T_{1}= \frac{16}{27}.




5 0
3 years ago
Solve for y in y+x=0
Naddika [18.5K]

Solve for y:

x + y = 0

Subtract x from both sides:

Answer:  y = -x

6 0
3 years ago
Please help <br>What is the area of this octagon?​
kolbaska11 [484]

Answer:

See the graphic I made

Area of Section A = 4 *1 = 4 square centimeters

Section B = 4 * (3 + 4) = 28 sq cm

Section C = 3 * (2 + 3 + 4) = 27 sq cm

Total Area = 4 + 28 + 27 = 59 square centimeters

Step-by-step explanation:

6 0
4 years ago
Why does the equation 35-5x+6x-36=-24-17x+17x+24 have one solution
seropon [69]
If it has only one solution that means that there is only one value of x that will fulfill this equation
35 - 5x + 6x - 36 = -24 - 17x + 17x + 24
-5x + 6x - 36 = -24 -17x + 17x + 24 - 35
x - 36 = -24 + 24 - 35 (I just canceled out the 17x because it would have justy made the process longer if I moved them over)
x = -24 + 24 - 35 + 36
x = -35 + 36
x = 1
3 0
3 years ago
Read 2 more answers
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