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3 years ago

Please help i'm struggling with this!

Mathematics

1 answer:

harkovskaia [24]3 years ago

6 0

Answer:

second option is correct (for question no 20)

Step-by-step explanation:

coordinate points are:

(3,1) = (x1,y1)

(5,9) = (x2,y2)

slope (m) = y2-y1/x2-x1

=9-1/5-3

=8/2

Now using slope intercept form

y=mx + b

y=4x + b

since the line passes through the point (3,1) it satisfies the equation

y=4x + b

1=4*3 + b

1-12 = b

-11 = b

substituting the value of b in equation

y=4x + b

y=4x +(-11)

y=4x - 11

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Find the value of x, circles and angles

lorasvet [3.4K]

Answer:

x = 50

Step-by-step explanation:

When two secants intersect in the interior of a circle, the angles formed are the average of the arc an angle and its vertical intercept. In this case, our angle, 73, should be the average of x and 96. We can translate this to an equation and solve:

$\frac{x+96}{2} = 73$

x + 96 = 146

x = 50

4 0

2 years ago

In the adjoining figure , APB and AQC are equilateral triangles. Prove that PC = BQ. ( Hint : <img src="https://tex.z-dn.net/?f=

just olya [345]

Answer:

See Below.

Step-by-step explanation:

Statements: Reasons:

$\displaystyle 1)\text{ } \Delta APB \text{ and } \Delta AQC \text{ are equilateral triangles}$ Given

$\displaystyle 2) \text{ } m \angle PAB = 60$ Definition of equilateral.

$3)\text{ } m \angle QAC = 60$ Definition of equilateral.

$4)\text{ } m\angle PAB = m\angle QAC$ Substitution

$5)\text{ } m\angle PAC=m\angle PAB+m\angle BAC$ Angle Addition

$\displaystyle 6)\text{ } m\angle QAB=m\angle QAC+m\angle BAC$ Angle Addition

$7)\text{ } m\angle QAB=m\angle PAB+m\angle BAC$ Substitution

$\displaystyle 8)\text{ } m\angle PAC=m\angle QAB$ Substitution

$9)\text{ } PA=BA$ Definition of equilateral

$10)\text{ } AC=AQ$ Definition of equilateral

$\displaystyle 11)\text{ } \Delta PAC \cong \Delta BAQ$ Side-Angle-Side Congruence*

$\displaystyle 12)\text{ } PC=BQ$ CPCTC

* SAS Congruence:

PA = BA

∠PAC = ∠QAB

AC = AQ

6 0

2 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,

$\rule{200}{3}$

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

∴ Hence, the required answer is 17/7.

6 0

2 years ago

Help fast and thankssss god bless

Wewaii [24]

Answer:

336-6=330

330÷6=55

55 Students in each bus

242-11=231

231÷7=33

Each book was 33 dollars

5+30+100=135

The smallest number is 5

18-3=15

15÷8=1.8

Each candy bar was 1.8 dollars

14-9=5

Her allowance is 5 dollars

4+20+60=84

The smallest number is 4

12-7=5

5×2=10

She had 10 in the begining

4 0

2 years ago

The fraction names thousandths, so there should

iren [92.7K]

Answer:

seven thousand

Step-by-step explanation:

6 0

3 years ago