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pishuonlain [190]
3 years ago
9

Find the product. (-2x^2)^3 * 3x

Mathematics
2 answers:
Lelechka [254]3 years ago
7 0

Answer:

-24\cdot x^{7}

Step-by-step explanation:

We have been given an expression (-2x^2)^3\cdot3x. We are asked to find the product of our given expression.

Using exponent property (a^b)^c=a^{b\cdot c}, we will get:

(-2x^2)^3\cdot 3x=-2^3\cdot x^{2\cdot 3}\cdot3x

(-2x^2)^3\cdot 3x=-8\cdot x^{6}\cdot 3x

(-2x^2)^3\cdot 3x=-8\cdot 3\cdot x^{6}\cdot x

(-2x^2)^3\cdot 3x=-24\cdot x^{6}\cdot x

Using exponent property a^b\cdot a^c=a^{b+c}, we will get:

(-2x^2)^3\cdot 3x=-24\cdot x^{6+1}

(-2x^2)^3\cdot 3x=-24\cdot x^{7}

Therefore, the product of our given expression would be -24\cdot x^{7}.

oee [108]3 years ago
3 0
I hope this helps you

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How many positive integers between 1000 and 9999 inclusive
bekas [8.4K]

Answer:

Step-by-step explanation:

a.

first number is  1000-1+9=1008

9)1000(1

    9

-------

     10

       9

    -----

       10

         9

       ----

         1

       ----

last number is 9999

9| 9999

  ---------

    1111 |0

    --------

9999=1008+(n-1)9

9999-1008=(n-1)9

n-1=8991/9=999

n=999+1=1000

b.

first digit=1000

last digit=9999-1=9998

2 |9999

  ---------

  |4999|1

9998=1000+(n-1)2

(n-1)2=9998-1000=8998

n-1=4499

n=4499=1=5000

c.not sure

d.

total  numbers=9000

9999=1000+(n-1)1

9999-1000=n-1

n=8999+1=9000

numbers divisible by 3=3000

first number=1002

last number=9999

9999=1002+(n-1)3

(n-1)3=9999-1002=8997

n-1=2999

n=2999+1=3000

numbers not divisible by 3=9000-3000=6000

e.

numbers divisible by 5=1800

first number=1000

last number=9995

9995=1000+(n-1)5

(n-1)5=9995-1000=8995

n-1=1799

n=1799+1=1800

numbers divisible by 7=1286

7 | 1000

  ---------

  |  142-6

1000-6+7=1001

7 | 9999

  |---------

    1428-3

9999-3=9996

first digit=1001

last digit=9996

9996=1001+(n-1)7

(n-1)7=9996-1001=8995

n-1=1285

n=1285+1=1286

numbers divisible by 35=257

first digit=1015

35 ) 1000 ( 28

        70

       ----

        300

        280

        ------

           20

           ---

1000-20+35=1015

35)9999(285

     70

    ----

     299

     280

     -----

        199

        175

        ----

          24

         ----

last digit=9999-24=9975

9975=1015+(n-1)35

(n-1)35=9975-1015=8960

n-1=8960/35=256

n=257

reqd. numbers=1800+1286-257=3019

7 0
3 years ago
Can someone help me please!!
Reika [66]
Answer:

X = 12

4x + 12 = 5x
subtract 4x on each side and you're left with x=12
3 0
3 years ago
If AABC = APQR, then<br> AB =
Gennadij [26K]

Answer:

AB = PQ

Step-by-step explanation:

ABC = PQR

The angles are equal and the segments are equal

for the angles

A = P

B = Q

C = R

For the segments

AB = PQ

BC = QR

AC = PR

3 0
3 years ago
Read 2 more answers
At which points does the graph of f(x)=2x−4 cross the x-axis and y-axis?
blagie [28]

crosses x-axis at (2, 0 ) and y-axis at (0, - 4 )

To find where the graph crosses the x and y axes ( intercepts )

• let x = 0, in the equation for y- intercept

• let y = 0, in the equation for x- intercept

x = 0 : y = 0 - 4 = - 4 ⇒ (0, - 4 )

y = 0 : 2x - 4 = 0 ⇒ 2x = 4 ⇒ x = 2 ⇒ (2, 0 )


6 0
3 years ago
I will send you a pic of the question... &lt;3​
RSB [31]

Answer:

kk send the pictures

Step-by-step explanation:

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6 0
3 years ago
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