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Nana76 [90]
3 years ago
5

Bett you guys can help meif you do tho you don’t have corona ✌

Mathematics
1 answer:
blsea [12.9K]3 years ago
7 0
Guess I have corona....................
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Help me solve this problem please
Simora [160]

Answer:

6

Step-by-step explanation:

8 0
3 years ago
Hello! Please help asap
ValentinkaMS [17]

Answer:

Hello! answer: 55

Step-by-step explanation:

55 + 35 = 90 HOPE THAT HELPS!

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3 years ago
The triangle shaped entrance to the cavern i 1 1/2 feet tall and 4 feet wide whats the area
oee [108]
A=1/2 b*h
So substitute it in 1/2*1/2*4
1/2*1/2= 1/4
1/4*4= 4/4= 1 so the entrance is 1 foot tall
6 0
3 years ago
Read 2 more answers
Alaina has 7 in her bank account and earns 6 each day . joel has 48 in his banck account and earn 8 earch day in how many days w
Elis [28]

Answer:

I think you wrote something wrong.

Step-by-step explanation:

If Alaina has $7 in her bank account and Joel had $48 in his Joel already has more then Alaina but I will attempt to teach you how to solve these types of problems.

for this example I'm giving Alaina 79 dollars.

First we know were starting with 79 dollars and increasing 6 dollars each day so the equation for this is; y = 6x + 79 that is Alaina's equation.

For Joel our equation will be; y = 8x + 48.

Since they already have 79 and 48 dollars in their bank we put those as our B in the equation cause they are our y-intercepts. (Y = mx + b) Next we know their increasing by 6 and 8 dollars a day so we make those our m. Once we have these equations we can plot them and see where the lines meet. (thats our answer) I plotted them on desmos graphing calculator.

They crossed at (15.5, 172) Which means it will take 16 days for Joel to catch up they will both have 172 dollars in the bank account on the 16th day.

Hope this helps!

8 0
3 years ago
A box contains 8 two-inch screws. Four have a Phillips head and 4 have a slotted head. In how many ways can 4 screws be chosen s
hammer [34]

Answer:

There are 6\cdot 6=36 different ways to choose 4 screws such that 2 have a Phillips head and 2 have a slotted head.

Step-by-step explanation:

If 4 screws must be chosen so that 2 have a Phillips head and 2 have a slotted head, then you have to choose 2 screws with a Phillips head from 4 screws with a Phillips head and 2 screws with a slotted head from 4 screws with a slotted head.

You can choose 2 screws with a Phillips head from 4 screws with a Phillips head in

C^4_2=\dfrac{4!}{2!(4-2)!}=\dfrac{4!}{2!\cdot2!}=\dfrac{1\cdot2\cdot3\cdot4}{1\cdot2\cdot1\cdot2}=6

different ways.

You can choose 2 screws with a slotted head from 4 screws with a slotted head in

C^4_2=\dfrac{4!}{2!(4-2)!}=\dfrac{4!}{2!\cdot2!}=\dfrac{1\cdot2\cdot3\cdot4}{1\cdot2\cdot1\cdot2}=6

different ways.

In total there are 6\cdot 6=36 different ways to choose 4 screws such that 2 have a Phillips head and 2 have a slotted head.

7 0
3 years ago
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