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3 years ago

Can someone tell me how to solve this and give me the answer please?

Mathematics

1 answer:

kotykmax [81]3 years ago

7 0

9h^2+42h+49 no common factors trial and error try first term and last term all sings are positive since last term and middle terms are possitive factor leading term and coeficient and try those 9=3*3 49=7*7 (3h+7)(3h+7) 9h^2+21h+21h+49 9h^2+42h+49 true (3h+7)(3h+7)=(3x+7)^2 answer is (3h+7)^2 answe ris ths second one

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Use SOH CAH TOA to find the length of KN

slega [8]

KN= 20.7

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7 0

1 year 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 <u>923.558 m²</u>

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, $A_{req}$ ≈ <u>923.558 m²</u>

Learn more about figures circumscribed by a circle here:

brainly.com/question/16478185

6 0

3 years ago

The dividend is a factor of 60. The divisor is a factor of 18. The quotient is a multiple of 3. Use the numbers to find one poss

Diano4ka-milaya [45]

Answer:

A possible solution is 60/2 = 30

Step-by-step explanation:

The given parameters are;

The dividend is a factor of 60

The divisor is a factor of 18

The quotient is a multiple of 3

The dividend is the amount divided

The dividend is the number that is dividing the dividend

The quotient is the result of the division operation

The given numbers to use are 60, 2, 30, and 18

Therefore one possible solution is 60/2 = 30

The dividend = 60 which is a factor of 60 (60 × 1 = 60)

The divisor = 2 which is a factor of 18 (2 × 9 = 18)

The quotient = 30 which is a multiple of 3 (3 × 10 = 30)

6 0

3 years ago

Please help! I just need the second question answered given the first is correct.

Nataliya [291]

Answer: 0.970 ??

Step-by-step explanation:

5 0

3 years ago

Holly has 56 peices of paper and 8 paper clips. If she clips the same amount of paper together with each paper clip, how many pe

dedylja [7]

Answer:

56/8 = 7 pieces of paper

Step-by-step explanation:

3 0

3 years ago