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3 years ago

Find three consecutive odd integers such that the sum of the smaller two is three times the largest increased by seven

Mathematics

2 answers:

Novosadov [1.4K]3 years ago

3 0

three consecutive odd integers are : -17,-15,-13

vovangra [49]3 years ago

3 0

Three consecutive odd integers are : -17,-15,-13
-17 + -15 = -32 = 3(-13) + 7 CHECKS!

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Complete the table of values for the function f(x)=-x^2+8x-2

Kipish [7]

Answer:

f(1) = 7

f(2) = 18

f(3) = 31

f(4) = 46

f(5) = 63

f(6) = 82

f(7) = 103

f(8) = 126

f(9) = 151

f(10) = 178

Step-by-step explanation:

f(1) = (-1)^2+8(1)-2 = 7

Continue plugging in values...

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2 years ago

Brennan purchased an almond tart costing $18.56. He gave the cashier $19.55. How much change did the cashier give back to Brenna

11111nata11111 [884]

Answer:

0.99.

Step-by-step explanation:

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3 years ago

If 13cos theta -5=0 find sin theta +cos theta / sin theta -cos theta

Ivahew [28]

Step-by-step explanation:

We have to find the value of (sinθ + cosθ)/(sinθ - cosθ), when 13 cosθ - 5 = 0.

$\red{\frak{Given}} \begin{cases} & \sf {13\ cos \theta\ -\ 5\ =\ 0\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \big\lgroup Can\ also\ be\ written\ as \big\rgroup} \\ & \sf {cos \theta\ =\ {\footnotesize{\dfrac{5}{13}}}} \end{cases}$

Here, we're asked to find out the value of (sinθ + cosθ)/(sinθ - cosθ), when 13 cosθ - 5 = 0. In order to find the solution we're gonna use trigonometric ratios to find the value of sinθ and cosθ. Let us consider, a right angled triangle, say PQR.

Where,

PQ = Opposite side
QR = Adjacent side
RP = Hypotenuse
∠Q = 90°
∠C = θ

As we know that, 13 cosθ - 5 = 0 which is stated in the question. So, it can also be written as cosθ = 5/13. As per the cosine ratio, we know that,

$\rightarrow {\underline{\boxed{\red{\sf{cos \theta\ =\ \dfrac{Adjacent\ side}{Hypotenuse}}}}}}$

Since, we know that,

cosθ = 5/13
QR (Adjacent side) = 5
RP (Hypotenuse) = 13

So, we will find the PQ (Opposite side) in order to estimate the value of sinθ. So, by using the Pythagoras Theorem, we will find the PQ.

Therefore,

$\red \bigstar {\underline{\underline{\pmb{\sf{According\ to\ Question:-}}}}}$

$\rule{200}{3}$

$\sf \dashrightarrow {(PQ)^2\ +\ (QR)^2\ =\ (RP)^2} \\ \\ \\ \sf \dashrightarrow {(PQ)^2\ +\ (5)^2\ =\ (13)^2} \\ \\ \\ \sf \dashrightarrow {(PQ)^2\ +\ 25\ =\ 169} \\ \\ \\ \sf \dashrightarrow {(PQ)^2\ =\ 169\ -\ 25} \\ \\ \\ \sf \dashrightarrow {(PQ)^2\ =\ 144} \\ \\ \\ \sf \dashrightarrow {PQ\ =\ \sqrt{144}} \\ \\ \\ \dashrightarrow {\underbrace{\boxed{\pink{\frak{PQ\ (Opposite\ side)\ =\ 12}}}}_{\sf \blue{\tiny{Required\ value}}}}$

∴ Hence, the value of PQ (Opposite side) is 12. Now, in order to determine it's value, we will use the sine ratio.

$\rightarrow {\underline{\boxed{\red{\sf{sin \theta\ =\ \dfrac{Opposite\ side}{Hypotenuse}}}}}}$

Where,

Opposite side = 12
Hypotenuse = 13

Therefore,

$\sf \rightarrow {sin \theta\ =\ \dfrac{12}{13}}$

Now, we have the values of sinθ and cosθ, that are 12/13 and 5/13 respectively. Now, finally we will find out the value of the following.

$\rightarrow {\underline{\boxed{\red{\sf{\dfrac{sin \theta\ +\ cos \theta}{sin \theta\ -\ cos \theta}}}}}}$

By substituting the values, we get,

$\rule{200}{3}$

$\sf \dashrightarrow {\dfrac{sin \theta\ +\ cos \theta}{sin \theta\ -\ cos \theta}\ =\ {\footnotesize{\dfrac{\Big( \dfrac{12}{13}\ +\ \dfrac{5}{13} \Big)}{\Big( \dfrac{12}{13}\ -\ \dfrac{5}{13} \Big)}}}} \\ \\ \\ \sf \dashrightarrow {\dfrac{sin \theta\ +\ cos \theta}{sin \theta\ -\ cos \theta}\ =\ {\footnotesize{\dfrac{\dfrac{17}{13}}{\dfrac{7}{13}}}}} \\ \\ \\ \sf \dashrightarrow {\dfrac{sin \theta\ +\ cos \theta}{sin \theta\ -\ cos \theta}\ =\ \dfrac{17}{13} \times \dfrac{13}{7}} \\ \\ \\ \sf \dashrightarrow {\dfrac{sin \theta\ +\ cos \theta}{sin \theta\ -\ cos \theta}\ =\ \dfrac{17}{\cancel{13}} \times \dfrac{\cancel{13}}{7}} \\ \\ \\ \dashrightarrow {\underbrace{\boxed{\pink{\frak{\dfrac{sin \theta\ +\ cos \theta}{sin \theta\ -\ cos \theta}\ =\ \dfrac{17}{7}}}}}_{\sf \blue{\tiny{Required\ value}}}}$

∴ Hence, the required answer is 17/7.

6 0

2 years ago

PLEASE HELP GEOMETRY!!!

aliina [53]

The answer would be A

8 0

3 years ago

Luke invited 20 more boys than girls to his birthday party. However, only ¾ of the boys and ⅔ of the girls came. Exactly 19 of L

sveta [45]

Answer:

Step-by-step explanation:

Let:

Number of girls invited = x

Number of boys = x +20

3/4 of boy showed up

2/3 of girls showed up

Exactly 19 did not show up

Therefore, we could say:

Fraction of boys that didn't show up = 1 - 3/4 = 1/4

Fraction of girls that didn't show up = 1 - 2/3 = 1/3

That is,

[1/4 *(x + 20)] + [1/3*x] = 19

[(x + 20) / 4] + x /3 = 19

(3(x + 20) + 4x) / 12 = 19

3x + 60 + 4x = 228

7x = 228 - 60

7x = 168

x = 168 / 7

x = 24

Number of girls, x = 24

Number of boys, x + 20 = 24 + 20 = 44

Total friends invited = (24 + 44) = 68

8 0

3 years ago