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3 years ago

Is (10, –6) a solution to this system of equations? ASAP PLS IM IN CLASS

Mathematics

1 answer:

Rus_ich [418]3 years ago

3 0

Answer:

Yes.

Step-by-step explanation:

(10, -6) are the values to substitute in for x and y, respectively.

11(10) + 19(-6) = 110 - 114 = - 14

3(10) + 8(-6) = 30 - 48 = -18

You might be interested in

Whats the original slope of y-7/12=5/8(x+9)

Vitek1552 [10]

1) Find the first derivative of f(x). 2) Plug x value of the indicated point into f '(x) to find the slope at x. 3) Plug x value into f(x) to find the y coordinate of the tangent point. 4) Combine the slope from step 2 and point from step 3 using the point-slope formula to find the equation for the tangent line.

y - 7/12 = 5/8(x + 9)

y - 7/12 = 5/8x + 45/8

+7/12 +7/12

Original Slope is y = 5/8x + 149/24

7 0

3 years ago

Read 2 more answers

Random Questions To Answer For Points.

DaniilM [7]

Answer:

1. someone who walk 42 feet in 6 seconds

2. At the Grocery mart the strawberries cost $1.49 per lb. At Baldwin Hills market the strawberries code $1.33 per lb

so the strawberries they would get is 3lb.

Step-by-step explanation:

8 0

3 years ago

Pls help ill give points

Dafna1 [17]

I hope you don't find the solution too messy. If there is anything you don't get. Let me know and I will explain it:

6 0

4 years ago

1. Dylan invested $47,000 in an account paying an interest rate of 4% compounded annually. Assuming no deposits or withdrawals a

Vaselesa [24]

Answer:

(1) 13.0 years

(2) $y^{3} =$ {-1728, 1728}

Step-by-step explanation:

(1)

Compound annually:

$Pe^{rt}$ = A

(47000) $e^{(0.04)(t)}$ = 79200

$e^{(0.04)(t)}$ = $\frac{79200}{47000}$

ln( $e^{(0.04)(t)}$ ) = ln( $\frac{79200}{47000}$ )

ln and e cancel out.

(0.04)(t) = ln( $\frac{79200}{47000}$ )

t = $\frac{ln(\frac{79200}{47000})}{0.04}$

t = 13.0 years

(2)

$x^{2} +y^{2} = 153\\y = -4x$

Substitute y with -4x.

$x^{2} + (-4x)^{2} = 153$

Solve for x.

x = {-3, 3}

Plug in x values into any equation to find y.

y = -4(-3) and y = -4(3)

y = {-12, 12}

$y^{3}$ = $-12^{3}$ = -1728

$y^{3}$ = $12^{3}$ = 1728

5 0

3 years ago

Read 2 more answers

Consider the transformation T: x = \frac{56}{65}u - \frac{33}{65}v, \ \ y = \frac{33}{65}u + \frac{56}{65}v

irina1246 [14]

Answer:

Step-by-step explanation:

$T: x = \frac{56}{65}u - \frac{33}{65}v, \ \ y = \frac{33}{65}u + \frac{56}{65}v$

$\frac{d(x,y)}{d(u,v)} =\left|\begin{array}{ccc}x_u&x_v\\y_u&y_v\end{array}\right|$

$=(\frac{56}{65} )^2+(\frac{33}{65} )^2\\\\=\frac{(56)^2+(33)^2}{(65)^2} \\\\=\frac{4225}{4225} \\\\=1$

B )

$S:-65 \leq u \leq 65, -65 \leq v \leq 65$

$T(65,65)=(x=\frac{56}{65} (65)-\frac{33}{65} (65),\ \ y =\frac{33}{65} (65)+\frac{56}{65} (65)\\\\=(23,89)$

$T(-65,65)=(-56-33,\ \ -33+56)\\\\=(-89,23)$

$T(-65,-65) = (-56+33,-33-56)\\\\=(-23,-89)$

$T(65,-65)=(56+33, 33-56)\\\\=(89,-23)$

$\int \!\! \int_{T(S)} \ x^2 + y^2 \ {dA}$

$=\int\limits^{65}_{v=-65} \int\limits^{65}_{u=-65}(x^2+y^2)(\frac{d(x,y)}{d(u,v)} du\ \ dv$

Now

$x^2+y^2=(\frac{56}{65} u-\frac{33}{65} v)^2+(\frac{33}{65} u+\frac{56}{65} v)^2$

$[(\frac{56}{65} )^2+(\frac{33}{65}) ^2]u^2+[(\frac{33}{65} )^2+(\frac{56}{65}) ^2]v^2$

$=\frac{(65)^2}{(65)^2} u^2+\frac{(65)^2}{(65)^2} v^2=u^2+v^2$

$\int \!\! \int_{T(S)} \ x^2 + y^2 \ {dA}$

$=\int\limits^{65}_{v=-65} \int\limits^{65}_{u=-65}(u^2+v^2) du\ \ dv$

$=\int\limits^{65}_{-65}\int\limits^{65}_{-65}u^2du \ \ dv+\int\limits^{65}_{-65}\int\limits^{65}_{-65}v^2du \ \ dv$

By symmetry of the region

$=4\int\limits^{65}_0 \int\limits^{65}_0u^2 du \ \ dv + u\int\limits^{65}_0 \int\limits^{65}_0v^2 du \ \ dv$

$= 4(\frac{u^3}{3} )^{65}_{0}(v)_0^{65}+(\frac{v^3}{3} )^{65}_{0}(u)_0^{65}\\\\=4[\frac{(65)^4}{3} +\frac{(65)^4}{3} ]$

$=\frac{8}{3} (65)^4$

8 0

3 years ago