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Gala2k [10]
3 years ago
9

Where does the graph of y = –3x – 18 intersect the x-axis?

Mathematics
1 answer:
Naya [18.7K]3 years ago
5 0

Answer:

b) (-6, 0)

Step-by-step explanation:

The quickest method to use is to put "0" in for <em>y</em><em>,</em><em> </em>then figure out which term for <em>x</em> would make the equation authentic, or genuine.

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1. Find the value of x.<br> X+ 15.92 = 21.378
OLEGan [10]

Answer:

x= 5.458

Step-by-step explanation:

x+15.92 = 21.378

  - 15.92   - 15.92

x = 5.458

Helpful hint: You just have to use inverse opperation to solve these problems

So, basically the opposite of what it is.

8 0
3 years ago
Sarris savings account balance change by $34 one week and by -$67 the next week which amount represents the greatest change
anygoal [31]
34 because it's a bigger number and -$67 is lower
8 0
4 years ago
D.
storchak [24]

Question:

Lisa walks at an average speed of  6 km/h for 15 hours.

Dan completes the same distance in  an hour.  Calculate Dan's average speed.​

Answer:

Speed = 90km/hr

Step-by-step explanation:

Given

Lisa:

Speed = 6km/hr

Time = 15\ hr

First, we calculate the distance covered by Lisa:

Distance = Speed * Time

Distance = 6km/hr * 15\ hr

Distance = 90km

So, Dan distance can be represented as:

Distance = 90km --- Dan completes the same distance as Lisa

Time = 1\ hr -- Time to complete the distance

Dan's average speed is:

Speed = \frac{Distance}{Time}

Speed = \frac{90km}{1\ hr}

Speed = 90km/hr

<em>Hence, Dan's average speed is 90km/hr</em>

4 0
3 years ago
Solve for p: p(5-2n) = 5/6-1/3
Hatshy [7]
Factor out P:
p(-3n+5)=½
Divide both sides by -3n+5:
p(-3n+5)^-3n+5

½^-3n+5

p=1^-6n+10
6 0
4 years ago
Find the ninth term of the sequence sqrt 5, sqrt 10, ^2sqrt 5
hjlf
The rule of geometric sequence is   ⇒⇒⇒ a * r^(n-1)
Where a is the first term and  r is the common ratio
for the given sequence  √5 , √10 , 2√5 , .......
a = √5
r = √10 / √5 = √2
The ninth term = √5 * (√2)^(9-1) = √5 * (√2)⁸ = 16 √5

the correct answer is the third option 16√5
5 0
4 years ago
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