Please help me out. I will mark brainlist. How do I solve this?

Mathematics

1 answer:

bulgar [2K]3 years ago

3 0

Okay so set up the equation to be

10b+45 = 6b+77

Okay Step 2: -6b from both sides and subtract 45 from both sides. Then your equation should look like...

2b = 32

Step 3: divide both sides by 2. Which should leave you with....

B=16

Now that we know B we have to plug it in so to solve for H the equation will be...

10(16)+45=205

That’s all!!

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If n is a term of the sequence 14, 8, 2, -4, …, which expression would you give the value of n?

Ipatiy [6.2K]

Answer:

-6n+20

Step-by-step explanation:

The given term of sequece is 14, 8, 2, -4, …,

Here the first term is 14 and then the terms are decreasing by 6 each time because $14-8=8-2=...=6$

As we know that the nth term of the sequence is given by the formula

$t_n=a+(n-1)d\\\text{Substitute the values we get}\\t_n=14+(n-1)(-6)\\t_n=14-6n+6\\t_n=20-6n\\$

If n is a term of the given sequence sequence, then its value is -6n+20

7 0

3 years ago

1. Find the equation of the line that is parallel to the line y = 0.6x-3 and passes through the point (-2,2).

AnnZ [28]

So the equation is parallel and the slope will be 2
y=0.6x+b
sub in(-2,2) for x and y to find b
2=0.6(-2)+b
2+1.2=b
b=3.2
therefore the equation is y=0.6x+3.2

8 0

3 years ago

7. Use the quadratic formula to find the solution(s). x² + 2x - 8 = 0

morpeh [17]

Hey there!

Use the quadratic formula to find the solution(s). x² + 2x - 8 = 0

Answer :

x = -4 or x = 2 ✅

Explanation :

Quadratic formula : ax² + bx + c = 0 where a ≠ 0

The number of real-number solutions (roots) is determined by the discriminant (b² - 4ac) :

If b² - 4ac > 0 , There are 2 real-number solutions

If b² - 4ac = 0 , There is 1 real-number solution.

If b² - 4ac < 0 , There is no real-number solution.

The roots of the equation are determined by the following calculation:

$x = \frac{ - b \pm \sqrt{ {b}^{2} - 4ac } }{2a}$

Here, we have :

a = 1
b = 2
c = -8

1) Calculate the discriminant :

b² - 4ac ⇔ 2² - 4(1)(-8) ⇔ 4 - (-32) ⇔ 36

b² - 4ac = 36 > 0 ; The equation admits two real-number solutions

2) Calculate the roots of the equation:

▪️ (1)

$x_1 = \frac{ - b - \sqrt{ {b}^{2} - 4ac} }{2a} \\ \\ x_1 = \frac{ - 2 - \sqrt{36} }{2(1) } \\ \\ x_1 = \frac{ - 2 - 6}{2} \\ \\ x_1 = \frac{ - 8}{2} \\ \\ \blue{\boxed{\red{x_1 = -4}}}$

▪️ (2)

$x_2 = \frac{ - b + \sqrt{ {b}^{2} - 4ac } }{2a} \\ \\ x_2 = \frac{ - 2 + \sqrt{36} }{2(1)} \\ \\ x_2 = \frac{ - 2 + 6}{2} \\ \\ x_2 = \frac{4}{2} \\ \\ \red{\boxed{\blue{x_2 = 2}}}$