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3 years ago

How i can answer this question?, NO LINKS, if you answer correctly i'll give u brainliest!

Mathematics

2 answers:

Jobisdone [24]3 years ago

8 0

6(4a-3b)

if you draw a factor tree that goes like:

1 24

2 12

... ... (and so on, you'll find 6 is the highest)

slavikrds [6]3 years ago

5 0

6•(4a-3b)

Steps:

Rewrite -18 as 3• 6
Rewrite 24 as 4 • 6
= 4 • 6a + 3• 6b

= 6(4a-3b)

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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

A flagpole cast a shadow 16 feet long at the same time a pole 9 feet high casts a shadow 6feet long what is the height in feet o

Doss [256]

The pole would be 24 feet high.
X/16=9/6
X=24

7 0

4 years ago

Neeeed help plssss solve just one to see how the way to solve others because I don’t understand the question ❤️ pls pls pleaaaas

spin [16.1K]

The descriminant is b^2-4ac so for 26 it would be -5^2 - 4 (1) (7). The quadratic formula is -B plus or minus the square root of b squared minus 4ac over 2a

8 0

3 years ago

Please, I need to know how this question is done. please

oksian1 [2.3K]

subsitute x = 6 into the lengths

3(6) -7

18 - 7

11cm

(6) + 5

11cm

therefore it is a square when x = 6 because the sides are all equal

8 0

3 years ago

5+7k>=-30 and -4k +4>20

Slav-nsk [51]

Answer:

$\large\boxed{-5\leq k$

Step-by-step explanation:

$5+7k\geq-30\ \wedge\ -4k+4>20\\\\5+7k\geq-30\qquad\text{subtract 5 from both sides}\\7k\geq-35\qquad\text{divide both sides by 7}\\\boxed{k\geq-5}\\\\-4k+4>20\qquad\text{subtract 4 from both sides}\\-4k>16\qquad\text{change the signs}\\4k$

7 0

3 years ago