Help please, I need it

Mathematics

1 answer:

Helen [10]3 years ago

5 0

Partial Answer:

For #10 the solutions are 2 and 5

Step-by-step explanation:

Solutions for an equation can be x-intercepts, or where it touches the x or horizontal line. The equation in #10 touches the x line at 2 and 5.

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X-5y=2 and 9x+8=15 solve by substitution

lozanna [386]

$x-5y=2$
Hence $x=2+5y$ (Equation I)
$9x+8=15$ (Equation II)
Substituting Equation I in Equation II
$9(2+5y)+8=15$
$18+45y+8=15$
$26+45y=15$
$45y=-11$
Hence $y= - \frac{11}{45}$

Substituting this in Equation in I to get X

$x-5(- \frac{11}{45} )=2$
$x+ \frac{11}{9}=2$
$x= 2- \frac{11}{9}$
$x= \frac{17}{9}$

7 0

3 years ago

Find the 15th term in the following<br> arithmetic sequence :<br> 1, 6, 11, 16,

Sunny_sXe [5.5K]

Answer:

Step-by-step explanation:

This arithmetic sequence starts at 1 and increases by 5 every time. The formula for this sequence could be expressed as A(n) = 1 + 5(n-1).

Since you're finding the 15th term, substitute n = 15 into the formula.

A(15) = 1 + 5((15)-1) = 1 + 5(14) = 1 + 70 = 71

6 0

3 years ago

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