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3 years ago

The ratio of meatballs to the total number of toppings on a pizza is 2:5. Can this be thought of as a fraction?

Mathematics

2 answers:

ladessa [460]3 years ago

3 0

Answer:

A. no because ratios can never be thought of as fravtions.

AleksandrR [38]3 years ago

3 0

C, yes because it is part to whole

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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

What value of x is in the solution set of <br> 3(x-4)>5x+2?

Harrizon [31]

Answer:

=x<-7

If they're asking for work, I'd be glad to show.

5 0

3 years ago

Read 2 more answers

PLS HELP<br> What is the area, in square of the parallelogram below?

Arlecino [84]

Answer:

35 square units

Step-by-step explanation:

Area of the parallelogram = 7*5 = 35 square units

6 0

3 years ago

What is the standard form of this complex number? (24-10i)-(43+18i)

Yuki888 [10]

Answer:

-19-28i

Step-by-step explanation:

Simplify and write the answer in standard form, a+bi.

6 0

3 years ago

If a tellemarketer made 200 phone calls and 35 customers signed up what is the relative frequency of getting a customer and not

Snezhnost [94]

Answer:

82.5% and 17.5%

Step-by-step explanation:

The telemarketer made 200 phone calls and only 35 people signed up. Therefore, we need to find the amount of people who didnt.

200 - 35 = 165

165 people did not sign up.

This meants that the frequency of getting a customer is 35/200 and not is 165/200. In percentages,

The relative frequency of not getting a customer is 82.5%

The relative frequency of getting a customer is 17.5%

6 0

3 years ago