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3 years ago

Simlify 2a^b x 3a^b, ^ = squared

Mathematics

2 answers:

ioda3 years ago

7 0

6a to the power 4 x b^

Temka [501]3 years ago

4 0

$2 a^{b} * 3a^{b}$
2(2^2)(3)(2^2)
2(2^2)(3)(2^2)
=96

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Need help here plss

goldenfox [79]

Answer:

see below

Step-by-step explanation:

Thankfully, these inequalities are already simplified, so all we have to do here is graph the inequalities as they are.

To graph y > 2x - 3, first draw the line y = 2x - 3. Make sure to make the line dashed because the sign is ">", which means that the points on the line do not belong to the solution set. Now, we need to figure out which side of the line to shade on. Since the sign is ">", that means that we will shade "above" the line, or on the left side of it. (This is the blue area in the image.)

Repeat a similar process to graph y < x + 1. As usual, first draw the line y = x + 1, and again, make it a dashed line since the sign is "<". And since the sign is "<", that means that we will shade "below" the line, or on the right side of it. (This is the red area in the image.)

The solution set to this system is the area where the two shaded areas overlap, which is the purple area in the image below. Hope this helps!

4 0

3 years ago

In the adjoining equilateral triangle PQR , X , Y and Z are the middle points of the sides PQ , QR and RP respectively. Prove th

olasank [31]

Answer:

See Below

Step-by-step explanation:

Statements: Reasons:

$1)\text{ } \Delta PQR\text{ is equilateral}$ Given

$2)\text{ }PQ=QR=RP$ Definition of Equilateral

$3)\text{ } X, Y, Z\text{ are the midpoints of } PQ, QR, RP$ Given

$4)\text{ } PX=XQ$ Definition of Midpoint

$5)PQ=PX+XQ$ Segment Addition

$6)\text{ } PQ=2XQ$ Substitution

$7)\text{ } QY=YR$ Definition of Midpoint

$8)\text{ }QR=QY+YR$ Segment Addition

$9)\text{ } QR=2QY$ Substitution

$10)\text{ } 2XQ=2QY$ Substitution

$11)\text{ } XQ=QY$ Division Property of Equality

$12)\text{ } XQ=PX=QY=YR$ Transitive Property

$13)\text{ } RZ=ZP$ Definition of Midpoint

$14)\text{ } RP=RZ+ZP$ Segment Addition

$15)\text{ } RP=2RZ$ Substitution

$16)\text{ } 2XQ=2RZ$ Substitution

$17)\text{ } XQ=RZ$ Substitution

$18)\text{ } XQ=PX=QY=YR=RZ=ZP$ Transitive Property

$19)\text{ } \angle Q\cong \angle R\cong \angle P$ Definition of Equilateral

$20)\text{ } \Delta XQY\cong \Delta YRQ \cong \Delta ZPX$ SAS Congruence

$21)\text{ } XY\cong YZ\cong ZX$ CPCTC

$22)\text{ } \Delta XYZ\text{ is equilateral}$ Equilateral Triangle Theorem

5 0

3 years ago

Read 2 more answers

I know the selected answer is correct but I'm not too sure how to get that answer.

Kryger [21]

$\tt{ Hey \: there , \: Mr.Panda \: ! }$ ;)

♨ $\large{ \tt{ E \: X \: P \: L \: A \: N \: A \: T \: I\: O \: N}}:$

⤻ Before solving the given question , you should know the answer of these questions :

✺How do you find the hypotenuse , perpendicular and base when the angle ( $\theta \: , \alpha \: ,\beta$ ) is given ?

⇾ The longest side , which is the opposite side of right angle is the hypotenuse ( h ). There are two other sides , the opposite and the adjacent. The naming of these sides depends upon which angle is involved. The opposite is the side opposite the angle involved and it is called the perpendicular ( p ) . The adjacent us the side next to the angle involved ( buy not the hypotenuse ) and it is called the base ( b ).

☄ $\large{ \tt{REMEMBER}} :$

$\bf{ \sin \theta = \frac{opposite}{hypotenuse} = \frac{perpendicular}{hypotenuse} }$

$\bf{ \cos\theta = \frac{adjacent}{hypotenuse} = \frac{base}{hypotenuse} }$

$\bf{ \tan \theta = \frac{opposite}{adjacent} = \frac{perpendicular}{base} }$

In the above cases , $\theta$ is taken as the angle of reference.

♪ Our Q/A part ends up here! Let's start solving the question :

❈ $\large{ \tt{GIVEN}} :$

Perpendicular ( p ) = ? , Hypotenuse ( h ) = 18 & base ( b ) = 16

✧ $\large{ \tt{TO \: FIND} : }$

Value of tan $\theta$

✎ $\large{ \tt{SOLUTION}} :$

Firstly , Finding the value of perpendicular ( p ) using Pythagoras theorem :

❃ $\boxed{ \sf{ {h}^{2} = {p}^{2} + {b}^{2} }}$ [ Pythagoras theorem ]

$\large{ ⇢ \sf{p}^{2} + {b}^{2} = {h}^{2} }$

$\large{⇢ \sf{ {p}^{2} = {h}^{2} - {b}^{2} }}$

$\large{ ⇢\sf{ {p}^{2} = {18}^{2} - {16}^{2} }}$

$\large{⇢ \sf{ {p}^{2} = 324 - 256}}$

$\large{⇢ \sf{ {p}^{2} = 68}}$

$\large{⇢ \sf{p = \sqrt{68}}}$

$\large{ ⇢\sf{p = \boxed{ \tt{2 \sqrt{17}}} }}$

Okey, We found out the perpendicular i.e $\tt{2 \sqrt{17}}$ . Now , We know :

❊ $\large{ \sf{ \tan \theta} = \frac{perpendicular}{base} }$

$\large {\tt{↬ \: tan \theta = \frac{2 \sqrt{17} }{16}}}$

$\large{ \tt{ ↬ tan \theta = \frac{ \cancel{2} \: \sqrt{17} }{ \cancel{16} \: \: 8} }}$

$\large{ \tt{ ↬ \boxed{ \tt{tan \theta = \frac{ \sqrt{17} }{8}}}}}$

⟿ $\boxed{ \boxed{ \tt{OUR\: FINAL \: ANSWER : \boxed{ \underline{ \bf{ \frac{ \sqrt{17} }{8}}}}}}}$

۵ Yay! We're done!

♕ $\large\tt{RULE \: OF \:SUCCESS }:$

Never lose hope & keep on working ! ✔

ツ Hope I helped!

☃ Have a wonderful day / evening! ☼

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3 0

3 years ago

Read 2 more answers

Evaluate: -1 2/5 ÷ 14/15

Sergio [31]

Answer:

Step-by-step explanation: Turn the mixed number into an improper fraction. -1 2/5 becomes -7/5. To divide fractions, you keep the first fraction, change the division sign to mulitplication and flip the second fraction. Then just multiply across the top and across the bottom. You will get 105/70. Reduce by dividing both numerator and denomination by 35 Result will be 3/2.

3 0

3 years ago

Original price:$89.00;markdown:33%

Anuta_ua [19.1K]

The answer is $59.99 =)

7 0

3 years ago