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2 years ago

6

Carris is making necklaces to sell at a local craft fair. She makes the necklaces by hand, so she uses a table to determine how

many beads to make for the necklaces.

2 answers:

Dahasolnce [82]2 years ago

4 0

The beads will make the necklace feel structured so i think 4 beads

sasho [114]2 years ago

3 0

Answer: 48

Step-by-step explanation:

Each is 8x? So 8x1, 8x2 so on, then you get 48!

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Is 5*9=4666666445554646444646

Step2247 [10]

Answer:

No, it is 45

Step-by-step explanation:

5*9=45

6 0

2 years ago

Please help on this math problem. :))

ratelena [41]

Answer:

$15.23$

Step-by-step explanation:

First, we can make an equation by adding all the ages together and dividing it by the sum of the frequencies.

$(13*7)+(14*12)+(15*18)+(16*9)+(17*5)+(18*4)+(19*2)$

Now simplify.

$(13*7)+(14*12)+(15*18)+(16*9)+(17*5)+(18*4)+(19*2) = 868$

We can divide this number by the sum of the frequencies.

$\frac{868}{7+12+18+9+5+4+2} \\ \\ \frac{868}{57} \\ \\ 15.23$

7 0

2 years ago

Read 2 more answers

A thermometer is taken from a room where the temperature is 23oC to the outdoors, where the temperature is −9oC. After one minut

Helga [31]

Answer:

$(a)\ 15\textdegree C\\(b)\ After\ 3\ minutes\ and\ 52.5\ Seconds\ from\ start$

Step-by-step explanation:

Temperature in the room $=23\textdegree C$

Decrease in the temperature(outdoors) $=23-15=8\textdegree C$ in $1\ minute$ .

Rate of decrease $=8\textdegree C/minute$

$Let\ after\ x\ minutes\ temperature\ is\ y\\y=23-8x\ \ \ \ y\geq -9\textdegree C$

(a) After $4\ more\ minutes\ total\ time\ is=1+4=5\ minutes$

$y=23-5\times 8\\y=-17\\$

But it is not possible as outdoor temperature is $-9\textdegree C$ so after $-9\textdegree C$ temperature decrease will not occur.

Hence after $4$ more minutes temperature= $-9\textdegree C$

(b) Let after $t\ minutes$ temperature is $-8\textdegree C$

Here $x=t,\ y=-8\textdegree C$

$-8=23-8x\\8x=23+8\\8x=31\\x=\frac{31}{8}\\x=3.875\ minutes=3\ minutes\ and\ 52.5\ seconds$

3 0

2 years ago

I have an interesting intergral that seems pretty challenging. After attempting to solve it for some time, I've come up empty ha

serious [3.7K]

${x}^{4} { {e}^{x} }^{3} = 0 \\ \Leftrightarrow {x}^{4} = 0 \: \vee \: { {e}^{x} }^{3} = 0 \\ \Leftrightarrow x = 0\: \vee \:x \in \emptyset \\ \Leftrightarrow x = 0$

3 0

2 years ago

What is the midpoint of (-3, -5) and (5, 3)

tensa zangetsu [6.8K]

Step-by-step explanation:

for a mid point of (-3,-5) and(5,3) is.

(X1,Y1)= (-3,-5)

(X2,Y2)=(5,3)

now,

midpoint=(x1+x2)/2, ,(y1+y2)/2

=(-3+5)/2, (-5+3)/2

=2/2, (-2)/2

=( 1, -1)

:- midpoint=(1,-1)

8 0

2 years ago