Learning Task 2. Fill in the largest unit digit to make them divisible

I need help please!:D

klasskru [66]

Answer:ion know

Step-by-step explanation:

5 0

2 years ago

In an arithmetic sequence {an}, if a1 = 5 and d = 3, the first 4 terms in the sequence are

noname [10]

A1 represents the first term

d represents the common difference.....and since the common difference is 3, u have to add 3 to each term to find the next term....because with an arithmetic sequence u add to find the next term, whereas, in a geometric sequence u multiply to find the next term

5...first term
5 + 3 = 8...2nd term
8 + 3 = 11...3rd term
11 + 3 = 14...4th term

ur answer is B

6 0

3 years ago

Read 2 more answers

I need help :/ i have five questions

IrinaK [193]

Answer:

Step-by-step explanation:

1. The given equation is:

$\frac{3}{5}{\times}\frac{4}{5}$

Multiplying both the terms, we get

$\frac{12}{25}$

Option A is correct.

2. The given equation is:

$2\frac{1}{3}{\times}\frac{3}{4}$

Converting the improper fraction into the mixed fraction,

$\frac{7}{3}{\times}\frac{3}{4}$

= $\frac{7}{4}$

Option C is correct.

3. The given equation is:

$\frac{3}{16}{\times}\frac{4}{21}$

Multiplying both the terms, we get

= $\frac{1}{28}$

4. The given equation is:

$3\frac{8}{9}{\times}4\frac{4}{5}$

Converting the improper fraction into the mixed fraction,

$\frac{35}{9}{\times}\frac{24}{5}$

= $\frac{56}{3}$

5. One package contains= $5\frac{1}{4}=\frac{21}{4}$ pounds of raisins.

$4\frac{2}{3}$ packages contains= $4\frac{2}{3}{\times}5\frac{1}{4}$ pounds of raisins.

= $\frac{14}{3}{\times}\frac{21}{4}$

= $\frac{49}{2}$

Therefore, the baker used $\frac{49}{2}$ pounds of raisins.

5 0

3 years ago

The sum of two numbers is 44. One number is 2 more than twice the other. Write a system of equations that defines this situation

hram777 [196]

System: x + y = 44; x = 2y + 2

Solve with substitution: x = 30; y = 14

x + y = 44
x = 2y + 2

Substitute 2y + 2 for x

2y + 2 + y = 44

3y + 2 = 44

Subtract 2 on both sides.

3y = 42

y = 14

Now substitute back in.

x + 14 = 44

Subtract 14 on both sides.

x = 30

8 0

2 years ago

Read 2 more answers

How many pounds of a 15% copper alloy must be mixed with 700lb of a 30% copper alloy to maybe a 25.5% copper alloy

lisov135 [29]

Answer:

$300\; \rm lb$ .

Step-by-step explanation:

Let $x$ represent the mass (in pounds) of that $15\%$ copper alloy required, such that the final mixture would contain $25.5\%$ copper by mass.

Consider: if $x$ pounds of that $15\%$ copper alloy is mixed with $700$ pounds that $30\%$ copper alloy, what would be the mass of copper in the mixture?

Mass of copper in $x$ pounds of that $15\%$ copper alloy: $(0.15\, x)\; \rm lb$ .
Mass of copper in $700$ pounds of that $30\%$ copper alloy: $700 \times 0.30 = 210\; \rm lb$ .