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Alex777 [14]
3 years ago
11

Evaluate 3z + 6 when z=4​

Mathematics
2 answers:
irina1246 [14]3 years ago
7 0

Answer:

3(4)+6

12+6

18

Step-by-step explanation:

Finger [1]3 years ago
6 0

Answer:

18

Step-by-step explanation:

If z = 4

3z + 6

= 3 ( 4 ) + 6

= 12 + 6

= 18

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.. A pizza restaurant charges $12.50 for a large cheese pizza plus $1.50 per topping. Write
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C= 1.50t * 12.50
The slope of the line is the cost of each topping. Because each individual topping is $1.50, you will multiply it by t (the number of toppings).
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3 years ago
Solve for X if 2x-5√X -3 = 0​
viktelen [127]

Answer:

x=9

Step-by-step explanation:

-5 square root x =-2+3

5x=4x^2 −12x+9

Simplify  25x-4{x}^{2}+12x-925x−4x 2 +12x−9  to  37x-4{x}^{2}-937x−4x 2-9

then...

(4x−1)(x−9)=0

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4 0
3 years ago
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Divide the following fractions<br><br> 3/4 ÷ 2/3<br><br> 1/2<br> 8/9<br> 9/8<br> 2
kifflom [539]

Answer:

9/8

Step-by-step explanation:

3/4 ÷ 2/3

= 3/4 x 3/2

= (3 x 3)/ (4 x 2)

= 9/8

6 0
4 years ago
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What is 3.00 in a fraction?
Alexeev081 [22]
It would be equal to 300/100
3 0
3 years ago
For any positive integer n, the sum of the first n positive integers equals n(n+1)2n(n+1)2. what is the sum of all the even inte
inn [45]

Answer:

The sum of even integers between 99 and 301 is 20200.

Step-by-step explanation:

To find : what is the sum of all the even integers between 99 and 301?

Solution : The even integers between 99 and 301  

100 would be least such integer and 300 would be greatest integer

So series form is 100,102,104,106,.......,300  

Now applying concepts of an arithmetic progression :

Where, First term,  a=100

Common difference,d=102-100=104-102=.....= 2

Last term, l=300

Now the relationship between a,d and I  

l=a+(n-1)\times d

where n is number of terms  

300=100+(n-1)\times 2

200=(n-1)\times 2

100=n-1

101=n

Now we have to find sum of these 101 terms  

Sum of n terms of an arithmetic progression is

S_n=\frac{n}{2}[2a+(n-1)d]

S_{101}=\frac{101}{2}[2(100)+(101-1)2]

S_{101}=\frac{101}{2}[200+(100)2]

S_{101}=\frac{101}{2}[400]

S_{101}=101\times200]

S_{101}=20200

Sum of all even integers between 99 and 301 is 20200.

4 0
3 years ago
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