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slega [8]
3 years ago
8

Taylor is buying a new flashlight. He can choose from red, blue, and silver flashlights that come in both a large size and a sma

ll size.
Which list shows all the possible flashlights that Taylor can choose?
A large, small

B. red, blue, silver

c. small red, small blue, large red, large silver

D small red, large red, small blue, large blue, small silver, large silver
Mathematics
1 answer:
Paladinen [302]3 years ago
4 0

Answer:

D

Step-by-step explanation:

It has all the flashlight colors and sizes.

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Help please! this is due tomorrow so please help out! will mark as brainiest!!
lesya692 [45]

Answer:

\large\boxed{\bold{Q1}\ -\dfrac{128}{9}}\\\boxed{\bold{Q2}\ 2^{25}}

Step-by-step explanation:

\bold{Q1}\\\\\dfrac{-2(3^2\cdot3^3)^2}{3^{12}\cdot2^{-6}}\qquad\text{use}\ a^n\cdot a^m=a^{n+m}\\\\=-\dfrac{2(3^{2+3})^2}{3^{12}\cdot2^{-6}}=-\dfrac{2(3^5)^2}{3^{12}\cdot2^{-6}}\qquad\text{use}\ (a^n)^m=a^{nm}\\\\=-\dfrac{2\cdot3^{5\cdot2}}{3^{12}\cdot2^{-6}}=-\dfrac{2\cdot3^{10}}{3^{12}\cdot2^{-6}}\qquad\text{use}\ \dfrac{a^n}{a^m}=a^{n-m}\\\\=-2^{1-(-6)}\cdot3^{10-12}=-2^{7}\cdot3^{-2}\qquad\text{use}\ a^{-n}=\dfrac{1}{a^n}\\\\=-2^7\cdot\dfrac{1}{3^2}=-\dfrac{2^7}{3^2}=-\dfrac{128}{9}

\bold{Q2}\\\\\dfrac{4^2}{4^{-7}}\cdot(2^4\cdot2^3)\cdot(3^{-2})^0\qquad\text{use}\ \dfrac{a^n}{a^m}=a^{n-m},\ a^n\cdot a^m=a^{n+m},\ a^0=1\\\\=4^{2-(-7)}\cdot2^{4+3}\cdot1=4^{9}\cdot2^7=(2^2)^{9}\cdot2^7\qquad\text{use}\ (a^n)^m=a^{nm}\\\\=2^{2\cdot9}\cdot2^7=2^{18}\cdot2^7\qquad\text{use}\ a^n\cdot a^m=a^{n+m}\\\\=2^{18+7}=2^{25}

5 0
3 years ago
The figure below shows a quadrilateral ABCD with diagonal BD bisecting angle ADC:
Neko [114]
The answer would be B) AD=DC, the reason why is because line AD is a reflection of line DC
6 0
3 years ago
Read 2 more answers
The area of the sector formed by the 110 degree central angle is 50 units squared. What is the circumference of the circle
Mama L [17]

Answer:

The circumference of the circle is C=45.346 \:units.

Step-by-step explanation:

A sector is the part of a circle enclosed by two radii of a circle and their intercepted arc. A pie-shaped part of a circle.

The area of a circle is given by A_{circle}=\pi r^2

The formula used to calculate the area of a sector of a circle is:

A_{sector}=\frac{central \:angle}{360} \cdot {Area \:of \:whole \:circle}\\\\A_{sector}=\frac{\theta}{360} \cdot \pi r^2

The circumference of a circle is the distance around the outside of the circle and its given by

C=2\pi r

We know the central angle \theta = 110º and the area of the sector 50 units squared.

First, we use the formula to calculate the area of a sector to find the radius.

50=\frac{110}{360} \cdot \pi r^2\\\\\frac{110}{360}\pi r^2=50\\\\\frac{11\pi }{36}r^2=50\\\\11\pi r^2=1800\\\\r^2=\frac{1800}{11\pi }\\\\\mathrm{For\:}x^2=f\left(a\right)\mathrm{\:the\:solutions\:are\:}x=\sqrt{f\left(a\right)},\:\:-\sqrt{f\left(a\right)}\\\\r=\sqrt{\frac{1800}{11\pi }},\:r=-\sqrt{\frac{1800}{11\pi }}

The radius can't be negative. Therefore,

r=\sqrt{\frac{1800}{11\pi }}\approx 7.217 \:units

Next, we apply the formula for the circumference of a circle.

C=2\pi (7.217)=45.346 \:units

7 0
3 years ago
8. Teo makes a necklace of x wooden beads at $0.50
VARVARA [1.3K]

Answer:

0.5x + 7.5 / (x + 6)

Step-by-step explanation:

Let x be the number of wooden beads

wooden bead = $0.50 each

glass bead = $1.25 each

since average cost of bead = (cost of wooden bead + cost of glass bead)/ total number of beads

equation is (0.5x + (6*1.25))/(x + 6)

The model equation is therefore 0.5x + 7.5 / (x + 6)

3 0
3 years ago
Tim is selling tickets to a school sporting event to raise money for his club. He put some extra money in his box before he bega
mihalych1998 [28]
The price of each ticket will be the slope found from the data.  

Slope=m=(y2-y1)/(x2-x1)=(177.5-146.25)/(18-13)

m=6.25  (so the cost of the tickets is $6.25)

Now we can use any data point to solve for b, or the y-intercept of the line:

y=6.25x+b, using (177.5, 18)

177.5=6.25(18)+b

b=$65.00d

So he started with $65.00
5 0
3 years ago
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