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4 years ago

Solve the system of equations using elimination.

Mathematics

1 answer:

tester [92]4 years ago

8 0

Answer:

y = 16 + 3/11

x = -76/11

Step-by-step explanation:

2y - 5x = -2

3y + 2x = 35

__________

(2y - 5x = -2)*3

(3y + 2x = 35)*2

__________

6y - 15x = -6

6y +4x = 70

__________

(6y - 15x = -6) - (6y +4x = 70)

15x - 4x = -6 -70

__________

11x = -76

x = -76/11

__________

3y +2(-76/11) = 35

3y = 35 + 152/11

3y = 13 + 35 + 9/11

3y = 48 + 9/ 11

y = 16 + 3/11

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What is the volume enclosed by the slanted prism in the diagram? A. 225 cm3 B. 270 cm3 C. 300 cm3 D. 540 cm3

Leokris [45]

Answer:

Step-by-step explanation:

The volume of a slanted prism is defined as the product between the area of its base and its height.

Remember that height is the distance from the top to the bottom, reaching that bottom with a perpendicular angle.

So, in this case, the height is 9 cm, and the area of the base is 25 square centimeter.

This means the volume is

$V=Bh=25cm^{2}(9cm)=225cm^{3}$

Therefore, the right answer is A.

4 0

4 years ago

the equation shows the radius of the algae,f(d), in mm,after d days: f(d)=7(1.06)^d the radius of the algae was approximately 13

Andrei [34K]

Answer:

$\begin{gathered} A)\text{ }Domain:\text{ \lparen0,11\rparen} \\ B)\text{ the y-intercept represents that in 0 days the diameter of the algae will be 7 mm.} \\ C)\text{ rate of change= 0.64. This means that on average the algae grows 0.64mm per day.} \end{gathered}$

Step-by-step explanation:

If the radius of the algae was approximately 13.29 mm, substitute f(d)=13.29 and solve for d to determine the domain:

$\begin{gathered} 13.29=7\left(1.06\right)^d \\ \frac{13.29}{7}=1.06^d \\ d=\frac{\text{ log\lparen13.29/7\rparen}}{\text{ log\lparen1.06\rparen}} \\ d=11 \\ \text{ Then, the domain:} \\ (0,11) \end{gathered}$

B. The y-intercept is obtained when d=0, hence;

$\begin{gathered} d=0 \\ f(0)=7(1.06)^0 \\ f(0)=7 \end{gathered}$

Therefore, the y-intercept represents that in 0 days the diameter of the algae will be 7 mm.

C. Now, to determine the rate of change of a function, use the following equation:

$\text{ rate of change=}\frac{change\text{ over y}}{change\text{ over x}}$

Then, determine the value of f(4):

$\begin{gathered} f(4)=7(1.06)^4 \\ f(4)=8.83 \end{gathered}$

Hence, the rate of change on that interval of the function:

$\begin{gathered} \text{ rate of change=}\frac{13.29-8.83}{11-4} \\ \text{ rate of change=0.64 } \end{gathered}$

This means that on average the algae grows 0.64mm per day.

8 0

1 year ago

The altitude of a plane as it begins to descend is shown by the linear equation a = 28,500 – 1800m, where a is the plane’s altit

mixer [17]

We substitute m=9 in a=28500-1800m

a = 28500 - 1800(9) = 28500 - 16200 = 12300

Answer: 12,300 feet

3 0

3 years ago

If E(X)=100, E(Y)=120, E(Z) = 130, Var(X) = 9, Var(Y) = 16, Var(Z) = 25, Cov(X, Y)= - 10 Cov(X,Z) = 12, and Cov(Y,Z) = 14, then

vredina [299]

Answer:

(1) -0.833

(2) 0.80

(3) 0.70

(4) 390

(5) 90

(7) 48

Step-by-step explanation:

Given:

E (X) = 100, E (Y) = 120, E (Z) = 130

Var (X) = 9, Var (Y) = 16, Var (Z) = 25

Cov (X, Y) = -10, Cov (X, Z) = 12, Cov (Y, Z) = 14

The formulas used for correlation is:

$Corr (A, B) = \frac{Cov (A, B)}{\sqrt{Var (A)\times Var(B)}} \\$

(1)

Compute the value of Corr (X, Y)-

$Corr (X, Y) = \frac{Cov (X, Y)}{\sqrt{Var (X)\times Var(Y)}} \\=\frac{-10}{\sqrt{9\times16}} \\=-0.833$

(2)

Compute the value of Corr (X, Z)-

$Corr (X, Z) = \frac{Cov (X, Z)}{\sqrt{Var (X)\times Var(Z)}} \\=\frac{12}{\sqrt{9\times25}} \\=0.80$

(3)

Compute the value of Corr (Y, Z)-

$Corr (Y, Z) = \frac{Cov (Y, Z)}{\sqrt{Var (Y)\times Var(Z)}} \\=\frac{14}{\sqrt{16\times25}} \\=0.70$

(4)

Compute the value of E (3X+4Y-3Z)-

$E(3X+4Y-3Z)=3E(X)+4E(Y)-3E(Z)\\=(3\times100)+(4\times120)-(3\times130)\\=390$

(5)

Compute the value of Var (3X-3Z)-

$Var (3X-3Z)=[(3)^{2}\times Var(X)]+[(-3)^{2}\times Var (Z)]+(2\times3\times-3\times Cov(X, Z)]\\=(9\times9)+(9\times25)-(18\times12)\\=90$

(6)

Compute the value of Var (3X+4Y-3Z)-

$Var (3X+4Y-3Z)=[(3)^{2}\times Var(X)]+[(4)^{2}\times Var(Y)]+[(-3)^{2}\times Var (Z)]+[(2\times3\times4\times Cov(X, Y)]+[(2\times3\times-3\times Cov(X, Z)]+[(2\times4\times-3\times Cov(Y, Z)]\\=(9\times9)+(16\times16)+(9\times25)+(24\times-10)-(18\times12)-(24\times14)\\=-230$

But this is not possible as variance is a square of terms.

(7)

Compute the value of Cov (3X, 2Y+3Z)-

$Cov(3X, 2Y+3Z)=Cov(3X,2Y)+Cov(3X, 3Z)\\=6Cov(X, Y)+9Cov(X,Z)\\=(6\times-10)+(9\times12)\\=48$

4 0

3 years ago

Graph each question y =2x + 0.5

Vikki [24]

Use y=mx+b It helped me so much

8 0

4 years ago