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3 years ago

6

HELP SOON!!!! Zahra wants the equation below to have an infinite number of solutions when the missing number is placed in the bo

x. Box (x minus 3) + 2 x = negative (x minus 5) + 4 Which number should she place in the box? –3 –1 1 3

2 answers:

snow_lady [41]3 years ago

6 0

Answer:

-3

Step-by-step explanation:

I got it correct on edge 2020.

marin [14]3 years ago

5 0

Answer:

Your answer is -3

Step-by-step explanation:

I had this on a test and i got it

plz mark brainliest

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jarptica [38.1K]

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3 years ago

What is the altitude of the equilateral triangle?

MAXImum [283]

It’s not going to be C because 10 is already in the question

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3 years ago

Could someone answer and explain these please? Thank you!

Oksana_A [137]

Answer 1:

It is given that the positive 2 digit number is 'x' with tens digit 't' and units digit 'u'.

So the two digit number x is expressed as,

$x=(10 \times t)+(1 \times u)$

$x=10t+u$

The two digit number 'y' is obtained by reversing the digits of x.

So, $y=(10 \times u)+(1 \times t)$

$y=10u+t$

Now, the value of x-y is expressed as:

$x-y=(10t+u)-(10u+t)$

$x-y=10t+u-10u-t$

$x-y=9t-9u$

$x-y=9(t-u)$

So, $9(t-u)$ is equivalent to (x-y).

Answer 2:

It is given that the sum of infinite geometric series with first term 'a' and common ratio r<1 = $\frac{a}{1-r}$

Since, the sum of the given infinite geometric series = 200

Therefore, $\frac{a}{1-r}=200$

Since, r=0.15 (given)

$\frac{a}{1-0.15}=200$

$\frac{a}{0.85}=200$

$a=0.85 \times 200$

a=170

The nth term of geometric series is given by $ar^{n-1}$ .

So, second term of the series = $ar^{2-1}$ = ar

Second term = $170 \times 0.15$

= 25.5

So, the second term of the geometric series is 25.5

Step-by-step explanation:

8 0

3 years ago

What is the simplified form of the following expression? 7(^3 square root 2x) - 3 (^3 square root 16x) -3 (^3 square root 8x)

Pachacha [2.7K]

Answer:

The third option listed: $\sqrt[3]{2x} -6\sqrt[3]{x}\\$

Step-by-step explanation:

We start by writing all the numerical factors inside the qubic roots in factor form (and if possible with exponent 3 so as to easily identify what can be extracted from the root):

$7\sqrt[3]{2x} -3\sqrt[3]{16x} -3\sqrt[3]{8x} =\\=7\sqrt[3]{2x} -3\sqrt[3]{2^32x} -3\sqrt[3]{2^3x} =\\=7\sqrt[3]{2x} -3*2\sqrt[3]{2x} -3*2\sqrt[3]{x}=\\=7\sqrt[3]{2x} -6\sqrt[3]{2x} -6\sqrt[3]{x}$

And now we combine all like terms (notice that the only two terms we can combine are the first two, which contain the exact same radical form:

$7\sqrt[3]{2x} -6\sqrt[3]{2x} -6\sqrt[3]{x}=\\=\sqrt[3]{2x} -6\sqrt[3]{x}$

Therefore this is the simplified radical expression: $\sqrt[3]{2x} -6\sqrt[3]{x}\\$

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3 years ago

9 squared ? 11 ? 8 ? 4 ? 1 = 60 The ? Can be +,-,x or d division

dexar [7]

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3 years ago