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Lynna [10]
3 years ago
14

Please someone help me please!I'm struggling and I'll give extra point's!

Mathematics
1 answer:
alukav5142 [94]3 years ago
3 0

Answer:

The x variable has an exponent of 2

Step-by-step explanation:

To be a linear equation, the highest power must be 1 on the variables

2x^2 +y =7 is quadratic since x^2 has a power of 2

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Please help fast!!!! Find the seventh term of the geometric sequence if a1=7 and r=3
statuscvo [17]

Answer:

15309 Im pretty sure

Step-by-step explanation:

a7=ar^7

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7 x 2187=15309

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2 years ago
Find the volume r:7 h:7
sweet-ann [11.9K]

Answer:

V = 1077.566 units²

Step-by-step explanation:

Assuming this is a cylinder, the formula is (area of base) × (height)

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= (π(7)²)(7)

= (153.938)(7)

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3 0
3 years ago
15 POINTS!!! (16*4^3*8^8)^5 Find the answer and show work how you got it
777dan777 [17]

Answer:

2^170

Step-by-step explanation:

(16 x 4^3 x 8^8)^ 5

- start by converting it all to a common base

(2^4 x 2^3 x 2^24)^5

- calculate the product, by adding all the powers

(2^34)^5

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5 0
3 years ago
Read 2 more answers
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7 0
3 years ago
The sum of squares of two consecutive odd numbers is 514 find the numbers ​
amid [387]

Correct Question : The sum of squares of two consecutive positive odd numbers is 514. Find the numbers.

\rule{130}1

Solution :

Let the two consecutive positive odd numbers be x and (x + 2)

\rule{130}1

☯ \underline{\boldsymbol{According\: to \:the\: Question\:now :}}

:\implies\sf x^2 + (x + 2)^{2} = 514 \\\\\\:\implies\sf x^2 + x^2 + 4x + 4 = 514\\\\\\:\implies\sf 2x^2 + 4x + 4 = 514 \\\\\\:\implies\sf 2x^2 + 4x = 514 - 4\\\\\\:\implies\sf 2x^2 + 4x = 510 \\\\\\:\implies\sf 2x^2 + 4x - 510 = 0\\\\\\:\implies\sf x^2 + 2x - 255 = 0 \:\:\:\:\:\Bigg\lgroup \bf{Dividing\:by\:2}\Bigg\rgroup\\\\\\:\implies\sf x^2 + 17x - 15x - 255 = 0\\\\\\:\implies\sf x(x + 17) - 15(x + 17) = 0\\\\\\:\implies\sf (x + 17) (x - 15)\\\\\\:\implies\sf x = - 17\:or\:x = 15\\\\\\:\implies\underline{\boxed {\sf x = 15}}\:\:\:\:\:\Bigg\lgroup \bf{Positive\:odd\: number}\Bigg\rgroup

When,

\bullet\: \textsf {x = } \textbf {15}

\bullet\: \sf x + 2 = 15 + 2 =\textbf{ 17}

\therefore\:\underline{\textsf{The required positive integer is \textbf{15 and 17}}}.

5 0
3 years ago
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