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3 years ago

32 + -3 • 4 – (–5) =

Mathematics

2 answers:

Anna71 [15]3 years ago

5 0

Answer:

121

Step-by-step explanation:

32+(-3)×4-(-5)

29×4-(-5)

116-(-5)

116+(+5)

121

OLga [1]3 years ago

3 0

Answer:

The correct answer would be 25

Step-by-step explanation:

32+(−3)(4)−(−5)

=32+−12−(−5)

=20−(−5)

=25

Hope this helps have an amazing day!!

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The dance committee of Pine Bluff Middle School earns \$72$72dollar sign, 72 from a bake sale and will earn \$4$4dollar sign, 4

Brut [27]

Answer:

The correct answer is 72 + 4 × x $\geq$ 400.

Step-by-step explanation:

The dance committee of Pine Bluff Middle School earns $72 from a bake sale and will earn $4 for each ticket they sell to the Spring Fling dance.

The dance will cost $400.

Let x number of tickets the committee could sell.

Total money earned is 72 + 4 × x.

The committee should have some left over money also after spending for the dance which costs $400.

Thus the inequality to determine the number of tickets the committee could sell to have money left over after they pay for this year's dance is given by,

72 + 4 × x $\geq$ 400.

3 0

3 years ago

Read 2 more answers

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,

$\rule{200}{3}$

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