Ask question

Ask question

All categories

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.

You might be interested in

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

2 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