MATHEMATICS HIGH SCHOOL

Find the side of a square, whose area is equal to the area of a rectangle with sides 6.4m and 2.5m. Also find the perimeter of the square.

Answers

Answer 1

Answer:

Answer:

side: 4 metres

perimeter: 16 metres

Step-by-step explanation:

Let's first find the area of this rectangle.

The area of a rectangle is denoted by A = lw, where l is the length and w is the width. Here, the length is l = 6.4 and the width is w = 2.5. Plug these in:

A = lw

A = 6.4 * 2.5 = 16 metres squared

We want to find the side of a square with area 16. Suppose the side length is x. The area of a square is denoted by A = x * x = x², so set this equal to 16:

x² = 16

x = √16 = 4

Thus, the side length is 4 metres.

The perimeter of a square is denoted by P = 4s, where s is the side length.

Here the side length is 4 metres, as we found, so:

P = 4s = 4 * 4 = 16

Hence the perimeter is 16 metres.

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I need this TONIGHT!!! Three years ago, Jolene bought $750 worth of stock in a software company. Since then the value of her purchase has been increasing at an average rate of 5 1/2% per year. How much is the stock worth now? (Round each money calculation you make to the nearest cent.) The stock is worth $ now.

Answers

Answer:

$A\approx\$880.68$

Step-by-step explanation:

So, we know that Jolene bought an initial $750.

We also know that the purchase is increasing at an average rate of 5 1/2 %or 5.5%. In other words, this is being compounded.

So, we can use the compound interest formula, which is:

$A=P(1+(r)/(n))^(nt)$

Where A is the total amount, P is the principal value, r is the rate and n is the number of times compounded per year, and t is the amount of years.

So, substitute 750 for P. 5 1/2% is the same as 5.5% or 0.055 (you move the decimal two places to the left and remove the percent symbol) so substitute this for r. Since it's increasing yearly, n is 1. So, our formula is:

$A=750(1+0.055)^t$

Add:

$A=750(1.055)^t$

Since the stock was bought 3 years ago, the value now is t=3. So, substitute 3 for t and evaluate:

$A=750(1.055)^3$

Evaluate. Use a calculator:

$A\approx\$880.68$

And we're done!

Formula: A = P(1 + r/n)^t

We have these variables:

P = 750

r/n = 0.055

t = 3

Substitute and simplify:

A = 750(1 + 0.055)^3

A = 750(1.055)^3

A = 880.68

Best of Luck!

If T(n) = 6n + 2 what is the 3rd term

Answers

To find the 3rd term, just replace the 'n' with 3.

Therefore:

t(3) = 6(3) + 2
t(3) = 18 + 2
t(3) = 20.

The 3rd term is 20.

T(n)=6n+2 T3=6(3)+2 T3=20

How many roots do the following equations have? -12x^2 - 25x+5 +x^3=0

Answers

Answer:

There are 3 roots of the given equation.

Step-by-step explanation:

Given the equation

$-12x^2-25x+5+x^3=0$

we have to tell the number of roots of the given equation.

As the number of roots for an equation is equal to degree.

The degree of a polynomial is the highest power of its monomials with non-zero coefficients.

Hence, number of roots is the highest power in the equation.

Now, the equation is $-12x^2-25x+5+x^3=0$

The highest power i.e degree of equation is 3.

hence, there are 3 roots of the given equation.

$-12x^2 - 25x + 5 + x^(3) = 0$
$x^(3) - 12x^(2) - 25x + 5 = 0$
$x = \sqrt[3]{((-b^(3))/(27a^(3)) + (bc)/(6a^(2)) - (d)/(2a)) + \sqrt{((-b^(3))/(27a^(3)) + (bc)/(6a^(2)) - (d)/(2a))^(2) + ((c)/(3a) - (b^(2))/(9a^(2)))^(3)}} + \sqrt[3]{((-b^(3))/(27a^(3)) + (bc)/(6a^(2)) - (d)/(2a)) - \sqrt{((-b^(3))/(27a^(3)) + (bc)/(6a^(2)) - (d)/(2a))^(2) + ((c)/(3a) - (b^(2))/(9a^(2)))^(3)}} - (b)/(3a)$
$x = \sqrt[3]{((-(-12)^(3))/(27(1)^(3)) + ((-12)(-25))/(6(1)^(2)) - (5)/(2(1))) + \sqrt{((-(-12)^(3))/(27(1)^(3)) + ((-12)(-25))/(6(1)^(2)) - (5)/(2(1)))^(2) + ((-25)/(3(1)) - (-(-25)^(2))/(9(1)^(2)))^(3)}} + \sqrt[3]{((-(-12)^(3))/(27(1)^(3))}} + ((-12)(-25))/(6(1)^(2)) - (5)/(2(1))) - \sqrt{((-(-12)^(3))/(27(1)^(3)) + ((-12)(-25))/(6(1)^(2)) - (5)/(2(1)))^(2) + ((-25)/(3(1)) - (-(-25)^(2))/(9(1)^(2)))^(3) - (-12)/(3(1))}}$
$x = \sqrt[3]{((-(-1728))/(27(1)) + (300)/(6(1)) - (5)/(2)) + \sqrt{((-(-1728))/(27(1)^(3)) + (300)/(6(1)) - (5)/(2))^(2) + ((-25)/(3(1)) - (144)/(9(1)))^(3)}}} + \sqrt[3]{((-(-1728))/(27(1)) + (300)/(6(1)) - (5)/(2)) - \sqrt{((-(-1728))/(27(1)^(3)) + (300)/(6(1)) - (5)/(2))^(2) + ((-25)/(3(1)) - (144)/(9(1)))^(3)}}} - (-12)/(3)$
$x = \sqrt[3]{((1728)/(27) + (300)/(6) - 2(1)/(2)) + \sqrt{((1728)/(27) + (300)/(6) - 2(1)/(2))^(2) + ((-25)/(3) - (144)/(9))^(3)}} + \sqrt[3]{((1728)/(27) + (300)/(6) - 2(1)/(2)) - \sqrt{((1728)/(27) + (300)/(6) - 2(1)/(2))^(2) + ((-25)/(3) - (144)/(9))^(3)}} - 4$
$x = \sqrt[3]{(64 + 50 - 2(1)/(2)) + \sqrt{(64 + 50 - 2(1)/(2))^(2) + (-8(1)/(3) - 16)^(3)}} + sqrt[3]{(64 + 50 - 2(1)/(2)) - \sqrt{(64 + 50 - 2(1)/(2))^(2) + (-8(1)/(3) - 16)^(3)}} - 4$
$x = \sqrt[3]{(114 - 2(1)/(2)) + \sqrt{(114 - 2(1)/(2))^(2) + (-24(1)/(3))^(3)}} + \sqrt[3]{(114 - 2(1)/(2)) - \sqrt{(114 - 2(1)/(2))^(2) + (-24(1)/(3))^(3)}} - 4$
$x = \sqrt[3]{(112(1)/(2)) + \sqrt{(112(1)/(2))^(2) - (24(1)/(3))^(3)}} - \sqrt[3]{(112(1)/(2)) + \sqrt{(112(1)/(2))^(2) - (24(1)/(3))^(3)}} - 4$
$x = \sqrt[3]{112(1)/(2) + √(12656.25 - 14408.037)} + \sqrt[3]{112(1)/(2) + √(12656.25 - 14408.037)} - 4$
$x = \sqrt[3]{112(1)/(2) + √(-1751.787)} + \sqrt[3]{112(1)/(2) - √(-1751.787)} - 4$
$x = \sqrt[3]{112(1)/(2) + 41.855i} + \sqrt[3]{112(1)/(2) - 41.855i} - 4$
$x = -4 + \sqrt[3]{112(1)/(2) + 41.855i} + \sqrt[3]{112(1)/(2) - 41.855i}$

How do I solve this problem? 6m+3(2m+5)+7

Answers

The final expression will be 12m + 22 .

Given,

6m+3(2m+5)+7

Here,

6m+3(2m+5)+7

To solve the above expression firstly open the brackets by multiplying 3 inside the bracket .

So,

6m + 6m + 15 + 7

Now add the the terms having similar variables ,

So,

6m and 3m will be added,

= 12m

Now add the constant terms,

15 + 7 = 22

Thus the final expression will be ,

12m + 22

Know more about algebra,

brainly.com/question/953809

#SPJ6

1. Combine like terms. So, combine the numbers with a variable. Then the numbers without variables. (Also known as simplifying the expression)

6m+3(2m+5)+7
12m+22

I can't figure this outthis question was on my exam and I got it completely wrong
(I have a graphing calculator capble of matrices so yo don't have to solve the system of equations by hand, use rref (reduced row echelon form))

values of a, b, and c and the equation of the graph of the parabola
y=ax^2+bx+c
such that is passes through the points
(2,-15)
(-5,-29)
(-3,5)

rewrite it in the form (x-h)^2=4P(y-k)

show all work
if I were to sub the points in I would ge
(2,-15): -15=4a+2b+c
(-5,-29): -29=25a-5b+c
(-3,5): 5=9a-3b+c

then solve for a, b and c
I don't know how to solve, please help
(if I don't undestand your answer, I will either report or ask you to explain more)

Answers

Suppose that equation of parabola is
y =ax² + bx + c

Since parabola passes through the point (2,−15) then
−15 = 4a + 2b + c

Since parabola passes through the point (-5,-29), then
−29 = 25a − 5b + c

Since parabola passes through the point (−3,−5), then
−5 = 9a − 3b + c

Thus, we obtained following system:
4a + 2b + c = −15
25a − 5b + c = −29
9a − 3b + c = −5

Solving it we get that
a = −2, b = −4, c = 1

Thus, equation of parabola is
y = −2x²− 4x + 1

____________________

Rewriting in the form of
(x - h)² = 4p(y - k)

i) -2x² - 4x + 1 = y

ii) -3x² - 7x = y - 11
(-3x² and -7x are isolated)

iii) -3x² - 7x - 49/36 = y - 1 - 49/36
(Adding -49/36 to both sides to get perfect square on LHS)

iv) -3(x² + 7/3x + 49/36) = y - 3
(Taking out -3 common from LHS)

v) -3(x + 7/6)² = y - 445/36

vi) (x + 7/6)² = -⅓(y - 445/36)
(Shifting -⅓ to RHS)

vii) (x + 1)² = 4(-1/12)(y - 445/36)
(Rewriting in the form of 4(-1/12) ; This is 4p)

So, after rewriting the equation would be -

(x + 7/6)² = 4(-⅛)(y - 445/36)

__________________

I hope this is what you wanted.

Regards,
Divyanka♪
__________________

Suppose that equation of parabola is
y =ax² + bx + c

Since parabola passes through the point (2,−15) then
−15 = 4a + 2b + c

Since parabola passes through the point (-5,-29), then
−29 = 25a − 5b + c

Since parabola passes through the point (−3,5), then
5 = 9a − 3b + c

Thus, we obtained following system:
4a + 2b + c = −15
25a − 5b + c = −29
9a − 3b + c = 5

Solving it we get that
a = −3, b = −7, c = 11

Thus, equation of parabola is
y = −3x²− 7x + 11

____________________

Rewriting in the form of
(x - h)² = 4p(y - k)

i) -3x² - 7x + 11 = y

ii) -3x² - 7x = y - 11
(-3x² and -7x are isolated)

iii) -3x² - 7x - 147/36 = y - 11 - 147/36
(Adding -147/36 to both sides to get perfect square on LHS)

iv) -3(x² + 7/3x + 49/36) = y - 543/36
(Taking out -3 common from LHS)

v) -3(x + 7/6)² = y - 181/12

vi) (x + 7/6)² = -⅓(y - 181/12)
(Shifting -⅓ to RHS)

vii) (x + 1)² = 4(-1/12)(y - 181/12)
(Rewriting in the form of 4(-1/12) ; This is 4p)

So, after rewriting the equation would be -

(x + 7/6)² = 4(-1/12)(y - 181/12)

_________________
I have corrected the answer and wrote again since the time to correct the answer from that account was expired.
- Divyanka

If a student spent a total of $83 on tickets ,how many tickets did he buy.? using this funtion below "c=24.50t 9.50

Answers

83=24.50(x)+9.50
83.00-9.50=73.50=24.50x
73.50/24.50=x
x=3 tickets
Your welcome =)

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