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3 years ago

Write a word problem that you can solve using multiples to estimate the quotient. Include a solution.

Mathematics

2 answers:

siniylev [52]3 years ago

7 0

Answer:

Let us consider,

The problem: 'Find two numbers between which the quotient of 50 ÷ 7 lies'.

Solution: We will solve this in two steps.

Step 1: Form a table of multiples of 7.

Counting number Multiple of 7

1 7

2 14

3 21

4 28

5 35

6 42

7 49

8 56

Step 2: Use the table to find the multiples closest to 50.

Since, 49 = 7 × 7 and 56 = 7 × 8

Thus, 50 lies between 49 and 56.

Then, the quotient is either 7 or 8.

As 50 is closest to 49.

Hence, the quotient of 50 ÷ 7 is 7.

Lilit [14]3 years ago

3 0

Answer:

Word problem based that can be solved using multiples to estimate the quotient:

Melissa made 326 cupcakes. She packed four cupcakes into each box. How many boxes of cupcakes did she pack? How many cupcakes were left unpacked?

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Jenna saves $3 for every $13 dollars she earns. Vanessa saves $6 for every $16 she earns. Is Jenna's ratio of money saved to mon

jeka94

We have been given that Jenna saves $3 for every $13 dollars she earns. Vanessa saves $6 for every $16 she earns.

We can compare the ratios of saved money to earned money by Jenna and Vanessa as:

$\frac{3}{13} =\frac{6}{16}$

Let us simplify our equation.

$0.2307692307692308=0.375$

We can see that $0.231\neq 0.375$ , therefore, Jenna's ratio of money saved to money earned is not equivalent to Vanessa's ratio of money saved to money earned.

5 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:

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