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3 years ago

9

16. At a grocery store, salted peanuts in the shell cost 30 cents per ounce. Is $5.00 enough money to buy 1 pound of peanuts? If

it is, what amount of money will be left over? UNIT 8 LESSON 6

2 answers:

IceJOKER [234]3 years ago

7 0

Yes. There is sixteen ounces in a pound. At thirty cents an ounce, you can buy a pound for $4.80. That also leaves $.20.

Andre45 [30]3 years ago

7 0

Yes, with .20 remaining.

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The following graph shows Barry's monthly phone bill and the number of minutes used. About how many minutes did Barry consume if

hram777 [196]

Answer:

About 27.5 minutes

Step-by-step explanation:

When looking at the graph, you can see that the red line intersects about 27.5 minutes with $55.

6 0

3 years ago

Which of the following sets are subspaces of R3 ?

Ratling [72]

Answer:

The following are the solution to the given points:

Step-by-step explanation:

for point A:

$\to A={(x,y,z)|3x+8y-5z=2} \\\\\to for(x_1, y_1, z_1),(x_2, y_2, z_2) \varepsilon A\\\\ a(x_1, y_1, z_1)+b(x_2, y_2, z_2) = (ax_1+bx_2,ay_1+by_2,az_1+bz_2)$

$=3(aX_l +bX_2) + 8(ay_1 + by_2) — 5(az_1+bz_2)\\\\=a(3X_l+8y_1- 5z_1)+b (3X_2+8y_2—5z_2)\\\\=2(a+b)$

The set A is not part of the subspace $R^3$

for point B:

$\to B={(x,y,z)|-4x-9y+7z=0}\\\\\to for(x_1,y_1,z_1),(x_2, y_2, z_2) \varepsilon B \\\\\to a(x_1, y_1, z_1)+b(x_2, y_2, z_2) = (ax_1+bx_2,ay_1+by_2,az_1+bz_2)$

$=-4(aX_l +bX_2) -9(ay_1 + by_2) +7(az_1+bz_2)\\\\=a(-4X_l-9y_1+7z_1)+b (-4X_2-9y_2+7z_2)\\\\=0$

$\to a(x_1,y_1,z_1)+b(x_2, y_2, z_2) \varepsilon B$

The set B is part of the subspace $R^3$

for point C: $\to C={(x,y,z)|x$

In this, the scalar multiplication can't behold

$\to for (-2,-1,2) \varepsilon C$

$\to -1(-2,-1,2)= (2,1,-1)$ ∉ C

this inequality is not hold

The set C is not a part of the subspace $R^3$

for point D:

$\to D={(-4,y,z)|\ y,\ z \ arbitrary \ numbers)$

The scalar multiplication s is not to hold

$\to for (-4, 1,2)\varepsilon D\\\\\to -1(-4,1,2) = (4,-1,-2)$ ∉ D

this is an inequality, which is not hold

The set D is not part of the subspace $R^3$

For point E:

$\to E= {(x,0,0)}|x \ is \ arbitrary) \\\\\to for (x_1,0 ,0) ,(x_{2},0 ,0) \varepsilon E \\\\\to a(x_1,0,0) +b(x_{2},0,0)= (ax_1+bx_2,0,0)\\$

The $x_1, x_2$ is the arbitrary, in which $ax_1+bx_2$ is arbitrary

$\to a(x_1,0,0)+b(x_2,0,0) \varepsilon E$

The set E is the part of the subspace $R^3$

For point F:

$\to F= {(-2x,-3x,-8x)}|x \ is \ arbitrary) \\\\\to for (-2x_1,-3x_1,-8x_1),(-2x_2,-3x_2,-8x_2)\varepsilon F \\\\\to a(-2x_1,-3x_1,-8x_1) +b(-2x_1,-3x_1,-8x_1)= (-2(ax_1+bx_2),-3(ax_1+bx_2),-8(ax_1+bx_2))$

The $x_1, x_2$ arbitrary so, they have $ax_1+bx_2$ as the arbitrary $\to a(-2x_1,-3x_1,-8x_1)+b(-2x_2,-3x_2,-8x_2) \varepsilon F$

The set F is the subspace of $R^3$

5 0

3 years ago

Help me out pleaseee

spayn [35]

Answer:

Option B is the correct choice.

Step-by-step explanation:

The graph is attached below.

We have to find the $x$ intercept meaning the value of the point on $x-axis$ when $y=0$

So we will put the value of zero $(0)$ instead of $y$ in our given equation.

So here we have solved it algebraically.

$y=\frac{3}{4}x-3$

Putting $y=0$

$0=\frac{3}{4}x-3$

$0=\frac{3x-12}{4}$

Multiplying $4$ both sides.

$0=3x-12$

Adding $12$ both sides.

$12=3x$

Dividing with $3$ both sides.

$x=\frac{12}{3}=4$

So the x-intercept of the given equation is $4$ which can be written as $(4,0)$ in terms of coordinates.

Option B $(4,0)$ is the correct choice.

8 0

3 years ago

The vertex of this parabola is (2,-1). When the y value is 0, the x value is 5. What is the coefficient of the squared term in t

Schach [20]

Vertex form: y = a(x - h)² + k
h = 2
k = -1
y = 0
x = 5

0 = a(5 - 2)² - 1
0 = 9a - 1
a = 1/9

5 0

3 years ago

Just 6-8 pls n thx.......

alex41 [277]

6 minus 8 equals -2.

5 0

3 years ago