All categories

3 years ago

3(7 + 4)2 − 14 ÷ 7. i need help and the 2 is squared

Mathematics

2 answers:

olga2289 [7]3 years ago

8 0

(this answer has been removed)

lorasvet [3.4K]3 years ago

3 0

Answer:

361

Step-by-step explanation:

To solve this you need to use the Order of Operations or PEMDAS.

The Order of Operations is Parentheses, Exponents, Multiplication, Division, Addition, and Subtraction.(PEMDAS)

So first you need to solve what's in the parentheses which is 7 + 4, this equals 11.

Then rewrite the problem as 3*11^2 - 14/7

Next you need to solve the exponents which is 11 squared. So 11 squared equals 121.

3*121 - 14/7

Now you need to multiply 3 by 121 since multiplication comes next in the order of operations. So 3*121 = 363

363 - 14/7, Next you need to divide 14 by 7 which is 2.

363 - 2, 363 subtracted by 2 is 361

You might be interested in

Aleksandra started studying how the number of branches in her tree grows over time. Every 1.5 years, the number of branches incr

vova2212 [387]

Answer:

N = 52 * (9/7)^(t/1.5)

Step-by-step explanation:

This problem can be modelated as an exponencial problem, using the formula:

N = Po * (1+r)^(t/1.5)

Where P is the final value, Po is the inicial value, r is the rate and t is the amount of time.

In our case, we have that N is the final number of branches after t years, Po = 52 branches, r = 2/7 and t is the number of years since the beginning (in the formula we divide by 1.5 because the rate is defined for 1.5 years)

Then, we have that:

N = 52 * (1 + 2/7)^(t/1.5)

N = 52 * (9/7)^(t/1.5)

7 0

3 years ago

A garden is shaped in the form of a regular heptagon (seven-sided), MNSRQPO. A circle with center T and radius 25m circumscribes

Alenkinab [10]

The relationship between the sides MN, MS, and MQ in the given regular heptagon is $\dfrac{1}{MN} = \dfrac{1}{MS} + \dfrac{1}{MQ}$

The area to be planted with flowers is approximately 923.558 m²

The reason the above value is correct is as follows;

The known parameters of the garden are;

The radius of the circle that circumscribes the heptagon, r = 25 m

The area left for the children playground = ΔMSQ

Required;

The area of the garden planted with flowers

Solution:

The area of an heptagon, is;

$A = \dfrac{7}{4} \cdot a^2 \cdot cot \left (\dfrac{180 ^{\circ}}{7} \right )$

The interior angle of an heptagon = 128.571°

The length of a side, S, is given as follows;

$\dfrac{s}{sin(180 - 128.571)} = \dfrac{25}{sin \left(\dfrac{128.571}{2} \right)}$

$s = \dfrac{25}{sin \left(\dfrac{128.571}{2} \right)} \times sin(180 - 128.571) \approx 21.69$

$The \ apothem \ a = 25 \times sin \left ( \dfrac{128.571}{2} \right) \approx 22.52$

The area of the heptagon MNSRQPO is therefore;

$A = \dfrac{7}{4} \times 22.52^2 \times cot \left (\dfrac{180 ^{\circ}}{7} \right ) \approx 1,842.94$

$MS = \sqrt{(21.69^2 + 21.69^2 - 2 \times 21.69 \times21.69\times cos(128.571^{\circ})) \approx 43.08$

By sine rule, we have

$\dfrac{21.69}{sin(\angle NSM)} = \dfrac{43.08}{sin(128.571 ^{\circ})}$

$sin(\angle NSM) =\dfrac{21.69}{43.08} \times sin(128.571 ^{\circ})$

$\angle NSM = arcsin \left(\dfrac{21.69}{43.08} \times sin(128.571 ^{\circ}) \right) \approx 23.18^{\circ}$

∠MSQ = 128.571 - 2*23.18 = 82.211

The area of triangle, MSQ, is given as follows;

$Area \ of \Delta MSQ = \dfrac{1}{2} \times 43.08^2 \times sin(82.211^{\circ}) \approx 919.382^{\circ}$

The area of the of the garden plated with flowers, $A_{req}$ , is given as follows;

$A_{req}$ = Area of heptagon MNSRQPO - Area of triangle ΔMSQ

Therefore;

$A_{req}$ = 1,842.94 - 919.382 ≈ 923.558

The area of the of the garden plated with flowers, $A_{req}$ ≈ 923.558 m²

Learn more about figures circumscribed by a circle here:

brainly.com/question/16478185

6 0

3 years ago

18. What is the solution of |2x+5|<19? -19 < x < 19 -12 -7 -24 < x < 14

elena-14-01-66 [18.8K]

Answer:

thats a tough one

Step-by-step explanation:

4 0

3 years ago

Theorem: the sum of any two even integers equals 4k for some integer k. "proof: suppose m and n are any two even integers. by de

icang [17]

Remark

The proof is only true if m and n are equal. Make it more general.

m = 2k

n = 2v

m + n = 2k + 2v = 2(k + v).

k and v can be equal but many times they are not. From that simple equation you cannot do anything for sure but divide by 2.

There are 4 combinations

m is divisible by 4 and n is not. The result will not be divisible by 4.

m is not divisible by 4 but n is. The result will not be divisible by 4.

But are divisible by 4 then the sum will be as well. Here's the really odd result

If both are even and not divisible by 4 then their sum is divisible by 4

3 0

3 years ago

What is the difference 1/9 - 5/6 a 13/18 b 1/18 c - 1/18 d -13/18

Murrr4er [49]

1/9 - 5/6

Convert both of them into denominators of 18.

2/18 - 15/18

Subtract the numerators and keep the denominator:

-13/18

3 0

3 years ago