All categories

3 years ago

Please help meeeeeeeeeee

Mathematics

1 answer:

Leto [7]3 years ago

8 0

$5x-2y+z=-1 \\ 2x+y+2z=6 \\ x-3y-z=-5$
firat we need to get two equation with two varibles let us work on x,y
so adding the first and last one will yield
$5x-2y+z=-1 \\ x-3y-z=-5 \\ -------- \\ 6x-5y=-6$
now we since we used the first and third we need to use the second to get a correct system.let us multiply the third by 2 then add the second and third
$x-2y-z=-5 \\ multiply2 \\ 2x-4y-2z=-10 \\ 2x+y+2z=6 \\ -------- \\ 4x-3y=-4$
now we have two equation with the variables x and y
$6x-5y=-6 \\ 4x-3y=-4$ you can solve it algebraically but you can see that the only solution possible is y=0 and x=-1 we have the values for x and y let us choose one of the three main equation and substitute to get z let us pick the first equation 5x-2y+z=-1-->5(-1)-2(0)+z=-1---->-5+z=-1-------->z=4
to make sure the system works let us check by substituting into the three equations
the first one will be 5x-2y+z=-1--->5(-1)-2(0)+4=-1---->-5+4=-1--->-1=-1 first equation holds
the second equation 3x+y+2z=6---->2(-1)+0+2(4)=6--->-2+8=-6--->-6=-6 second equation holds
the third equation x-3y-z=-5----->-1-3(0)-4=-5---->-1-4=-5--->-5=-5
our third equation also holds which makes our solution correct
x=-1,y=0,z=4

You might be interested in

Two trains 100 km apart are traveling toward each other along the same track. The first train goes 50 km per hour; the second tr

dem82 [27]

Answer:

92.31 km

Step-by-step explanation:

Combined speed = 50+80

= 130 km/hour

Time = 100/130 = 10/13 hours to collide

Speed × time = distance

distance = 10/13 × 120

= 92.30769231 km

7 0

3 years ago

Y=3x+2<br><br> please help me lolz

Elanso [62]

Answer:

Slope (M): 3

Y-Int: 2

Standard Form: -3x+y=2

can go through points: (5,1), (-1, -1),...

3 0

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

3 years ago

Find x using the picture below.

Katen [24]

Answer:

x ≈ 14.5

Step-by-step explanation:

A = x²√(25+10√5)/4

A = 1.72a²

x² = A/1.72

= 363.27/1.72

≈ 211

x = √211

≈ 14.5

6 0

3 years ago

hey! please help with this!!! Write all the 12 answer of measurements. If you can’t do it and you think it’s a lot you don’t hav

Minchanka [31]

Answer:

Give it a bit of time OK I will say the answer is in the *c'h_a-t

7 0

3 years ago