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3 years ago

Please help with this math question

Mathematics

1 answer:

blagie [28]3 years ago

8 0

Answer:

Mirandah added 5x and 9y, which she shouldn't do because one is y and the other is x, meaning they don't match.

There is no simplifying the equation. It should stay 5x + 9y - 4

If wrong, I apologize, but this should be correct.

You might be interested in

how do you have a reflection in the x axis, if the shape is already partially in the third and second quadrent of the graph

IgorLugansk [536]

Its a reflection over an x-axis, whats "positive" goes "negative" and whats "negative" goes "positive" with respect to y.

If a shape $S$ is a subset (made of $P_2\cup P_3$ of quadrants $Q_2,Q_3$ then the part $P_2$ that lies in the second quadrant gets mapped to $Q_3$ and part $P_3$ that lies in third quadrant gets mapped to $Q_2$

Hope this helps.

8 0

3 years ago

He nine ring wraiths want to fly from barad-dur to rivendell. rivendell is directly north of barad-dur. the dark tower reports t

kiruha [24]

This problem is better understood with a given figure. Assuming that the flight is in a perfect northwest direction such that the angle is 45°, therefore I believe I have the correct figure to simulate the situation (see attached).

Now we are asked to find for the value of the hypotenuse (flight speed) given the angle and the side opposite to the angle. In this case, we use the sin function:

sin θ = opposite side / hypotenuse

sin 45 = 68 miles per hr / flight

flight = 68 miles per hr / sin 45

<span>flight = 96.17 miles per hr</span>

6 0

3 years ago

1+2-3+4+5-6+7+8-9...+97+98-99

zaharov [31]

Answer:

1584

Step-by-step explanation:

The sum of this sequence can be found a number of ways. One way is to recast it as the series whose terms are groups of three terms of the given series.

<h3>series of partial sums</h3>

The partial sums, taken 3 terms at a time, are

1+2-3 = 0

4+5-6 = 3

7+8-9 = 6

...

97+98-99 = 96

So the original series is equivalent to ...

0 +3 +6 +... +96 = 3×1 +3×2 +... +3×32 = 3×(1 +2 +... +32)

That is, the sum is 3 times the sum of the consecutive integers 1..32.

<h3>consecutive integers</h3>

The sum of integers 1..n is given by the equation ...

s(n) = n(n+1)/2

<h3>series sum</h3>

Using this to find the sum of our series, we find it to be ...

series sum = 3 × (32)(33)/2 = 1584

_____

<em>Alternate solution</em>

The given series is the sum of integers 1-99, with 6 times the sum of integers 1-33 subtracted. That is, ...

1 + 2 - 3 + 4 + 5 - 6 = 1+2+3+4+5+6 -2(3 +6) = 1+2+3+4+5+6 -6(1+2)

Continuing on to ...97 +98 -99 gives the result s(99) -6s(33).

Computed that way, we find the sum to be ...

(99)(100)/2 -6(33)(34)/2 = 4950 -3366 = 1584

3 0

2 years ago

Write an equivalent expression for each question

Mandarinka [93]

Answer:

A. 16 - 4 x = 4(4 - x)

B. 4 x + 8 = 4( x + 2)

C. -8 b - 24 = -8(b + 3)

D. 54 + 9 x = 9 (x + 6)

Step-by-step explanation:

Here, the given expressions are:

A. 16 - 4 x

Now, 16 and 4 have 4 as their COMMON FACTOR.

⇒Taking out 4 as the factor, we get: 16 - 4 x = 4 ( 4 - x)

Hence, the equivalent expression of 16 - 4 x = 4 ( 4 - x)

B. 4x + 8

Now, 4 and 8 have 4 as their COMMON FACTOR.

⇒Taking out 4 as the factor, we get: 4 x + 8 = 4 ( x + 2)

Hence, the equivalent expression of 4 x + 8 = 4 ( x + 2)

C. -8b - 24

Now, -8 and -24 have (-8) as their COMMON FACTOR.

⇒Taking out (-8) as the factor, we get: -8b - 24 = (-8) ( b+ 3)

Hence, the equivalent expression of -8b - 24 = (-8) ( b+ 3)

D. 54 + 9x

Now, 54 and 9 have (9) as their COMMON FACTOR.

⇒Taking out (9) as the factor, we get: 54 + 9 x = 9( 6 + x)

Hence, the equivalent expression of 54 + 9 x = 9( 6 + x)

4 0

3 years ago

David’s bank offers a 36-month Certificate of Deposit (CD) with an APR of 2.25%. Use the compound interest formula to answer the

tamaranim1 [39]

Using compound interest, it is found that:

a) A(8) = 2389.66

b) t = 31.15

c) P = 1870.85

Compound interest:

$A(t) = P\left(1 + \frac{r}{n}\right)^{nt}$

A(t) is the amount of money after t years.
P is the principal(the initial sum of money).
r is the interest rate(as a decimal value).
n is the number of times that interest is compounded per year.
t is the time in years for which the money is invested or borrowed.

In this problem:

The APR is of 2.25%, hence $r = 0.0225$ .

No information about the number of compounding per year, hence $n = 1$ .

Item a:

$P = 2000$ , hence:

$A(t) = P\left(1 + \frac{r}{n}\right)^{nt}$

$A(8) = 2000\left(1 + \frac{0.0225}{1}\right)^{8}$

$A(8) = 2389.66$

Item b:

$A(t) = 4000$ , hence:

$A(t) = P\left(1 + \frac{r}{n}\right)^{nt}$

$4000 = 2000\left(1 + \frac{0.0225}{1}\right)^{t}$

$(1.0225)^t = 2$

$\log{(1.0225)^t} = \log{2}$

$t\log{1.0225} = \log{2}$

$t = \frac{\log{2}}{\log{1.0225}}$

$t = 31.15$

Item c:

$A(3) = 2000$ , hence:

$A(t) = P\left(1 + \frac{r}{n}\right)^{nt}$

$2000 = P\left(1 + \frac{0.0225}{1}\right)^{3}$

$P = \frac{2000}{(1.0225)^3}$

$P = 1870.85$

A similar problem is given at brainly.com/question/24850750

7 0

2 years ago