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3 years ago

I need help with this :(

Mathematics

2 answers:

11111nata11111 [884]3 years ago

8 0

Answer:

The sum of the areas of square N and square L is equal to the area of square K

fiasKO [112]3 years ago

4 0

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51x+x+33x+10 I need help on combining like terms

Serhud [2]

Answer:

84x+10

Step-by-step explanation:

Ok, since 51x, x, and 33x all have x in it, you add them all together. From that, you should get 84x. We can not add 10 with anything, the expression in simplest form is 84x+10.

5 0

3 years ago

BRIA is a rectangle. AN=5. NR=x-3. please solve for x, NR, and BI

Lena [83]

Answer:

x=8

NR=5 units

BI=10 units

Step-by-step explanation:

In a rectangle BRIA

AN=5 units

NR=x-3

We have to solve for x , NR and BI.

We know that

Diagonals of rectangle bisect to each other.

BI and AR are the diagonals of rectangle BRIA and intersect at point N.

AN=NR

$5=x-3$

$x=5+3=8$

$x=8$

Substitute the value of x

NR=8-3=5

By property of rectangle

BI=AR=AN+NR=5+5=10 unit

BI=10 units

7 0

3 years ago

help with this plzzzz, i need perimeter and area plzzzz steps if possible will mark brainlest!!!!!!!!!!!!!!!!!!!!!

weqwewe [10]

Answer:

Area: 125

Perimeter: 60

Step-by-step explanation:

I got this answer by calculating the area and perimeter. To find the area, multiply length x width. To find the perimeter, you have to add your length and width two times.

3 0

3 years ago

Find the number of ways to distribute six different toys to three different children such that each child gets at least one toy.

Lisa [10]

Answer:

Step-by-step explanation:

Given that there are six different toys and they are to be distributed to three different children.

The restraint here is each child gets atleast one toy.

Let us consider the situation as this.

Since each child has to get atleast one toy no of ways to distribute

any 3 toys to the three children each. This can be done by selecting 3 toys from 6 in 6C3 ways and distributing in 3! ways

So 3 toys to each one in 6x5x4 =120 ways

Now remaining 3 toys can be given to any child.

Hence remaining 3 toys can be distributed in 3x3x3 =27 ways

Total no of ways

= 120(27)

= 3240

8 0

3 years ago

You deposit 2000 in account A, which pays 2.25% annual interest compounded monthly. You deposit another 2000 in account b, which

stellarik [79]

To model this situation, we are going to use the compound interest formula: $A=P(1+ \frac{r}{n} )^{nt}$
where
$A$ is the final amount after $t$ years
$P$ is the initial deposit
$r$ is the interest rate in decimal form
$n$ is the number of times the interest is compounded per year
$t$ is the time in years

For account A:
We know for our problem that $P=2000$ and $r= \frac{2.25}{100} =0.0225$ . Since the interest is compounded monthly, it is compounded 12 times per year; therefore, $n=12$ . Lets replace those values in our formula:
$A=2000(1+ \frac{0.0225}{12} )^{12t}$

For account B:
$P=2000$ , $r= \frac{3}{100} =0.03$ , $n=12$ . Lest replace those values in our formula:
$A=2000(1+ \frac{0.03}{12} )^{12t}$

Since we want to find the time, $t$ , when the sum of the balance in both accounts is at least 5000, we need to add both accounts and set that sum equal to 5000:
 $2000(1+ \frac{0.0225}{12} )^{12t}+2000(1+ \frac{0.03}{12} )^{12t}=5000$

Now that we have our equation, we just need to solve for $t$ :
$2000[(1+ \frac{0.0225}{12} )^{12t}+(1+ \frac{0.03}{12} )^{12t}]=5000$
$(1+ \frac{0.0225}{12} )^{12t}+(1+ \frac{0.03}{12} )^{12t}= \frac{5000} {2000}$
$(1.001875)^{12t}+(1.0025 )^{12t}= \frac{5}
{2}$
$ln(1.001875)^{12t}+ln(1.0025 )^{12t}=ln( \frac{5} {2})$
$12tln(1.001875)+12tln(1.0025 )=ln( \frac{5} {2})$
$t[12ln(1.001875)+12ln(1.0025 )]=ln( \frac{5} {2})$
$t= \frac{ln( \frac{5}{2} )}{12ln(1.001875)+12ln(1.0025 )}$
$17.47$

We can conclude that after 17.47 years the sum of the balance in both accounts will be at least 5000.

5 0

3 years ago