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2 years ago

Help me with this one pls

Mathematics

2 answers:

vlada-n [284]2 years ago

7 0

Answer:

look at the picture attached

bagirrra123 [75]2 years ago

6 0

1/3y + 28 = -5

Subtract 28 from both sides:

1/3y = -33

Multiply both sides by 3:

y = -99

The answer is E) NOTA

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Marco borrowed $150 from his brother.He has paid him back 30% so far.How much money does Marco still owe his brother?

alexdok [17]

150*30% = $45
$150-$45 = $105

Marco still owes his brother $105

8 0

3 years ago

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The eleventh term from the end of the AP 14, 19, 24, …………., 264 is

vazorg [7]

Answer:

<h2>The eleventh term of the sequence is 64</h2>

Step-by-step explanation:

The sequence given is an arithmetic sequence

14, 19, 24, …………., 264

The nth term of an arithmetic sequence is given as;

Tn = a+(n-1)d where;

a is the first term = 14

d is the common difference = 19-14=24-19 = 5

n is the number of terms = 11(since we are to look for the eleventh term of the sequence)

substituting the given values in the formula given;

T11 = 14+(11-1)*5

T11 = 14+10(5)

T11 = 14+50

T11 = 64

The eleventh term of the sequence is 64

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3 years ago

A sandwich shop is 70 stores and 70% of the stores are in California the rest of stores are in Nevada how many stars are in Cali

sineoko [7]

49 stores in California
21 stores in Nevada

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3 years ago

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The sum of the polynomials 6x3 + 8x2 – 2x + 4 and 10x3 + x2 + 11x + 9 is

faltersainse [42]

Answer:

16x^2 + 9x^2 + 9x + 13.

Step-by-step explanation:

6x^3 + 8x^2 – 2x + 4 +10x^3 + x^2 + 11x + 9

Bringing like terms together:

= 6x^3 + 10x^3 + 8x^2 + x^2 - 2x + 11x + 4 + 9

= 16x^2 + 9x^2 + 9x + 13. (answer).

4 0

3 years ago

The set of points (-3, 4), (-1, 1), (-3,-2), and (-5,1) identifies the vertices of a quadrilateral. Which is the most specific d

KIM [24]

Answer:

Option (4). Rhombus

Step-by-step explanation:

From the figure attached,

Distance AB = $\sqrt{(x_{2}-x_{1})^2+(y_{2}-y_{1})^2}$

= $\sqrt{(1-4)^2+(-5+3)^2}$

= $\sqrt{(-3)^2+(-2)^2}$

= $\sqrt{13}$

Distance BC = $\sqrt{(4-1)^2+(-3+1)^2}$

= $\sqrt{9+4}$

= $\sqrt{13}$

Distance CD = $\sqrt{(-2-1)^2+(-3+1)^2}$

= $\sqrt{9+4}$

= $\sqrt{13}$

Distance AD = $\sqrt{(1+2)^2+(-5+3)^2}$

= $\sqrt{9+4}$

= $\sqrt{13}$

Slope of AB ( $m_{1}$ ) = $\frac{y_{2}-y_{1}}{x_{2}-x_{1}}$

= $\frac{4-1}{-3+5}$

= $\frac{3}{2}$

Slope of BC ( $m_{2}$ ) = $\frac{4-1}{-3+1}$

= $-\frac{3}{2}$

If AB and BC are perpendicular then,

$m_{1}\times m_{2}=-1$

But it's not true.

[ $m_{1}\times m_{2}=(\frac{3}{2})(-\frac{3}{2})$ = - $\frac{9}{4}$ ]

It shows that the consecutive sides of the quadrilateral are not perpendicular.

Therefore, ABCD is neither square nor a rectangle.

Slope of diagonal BD = $\frac{4+2}{-3+3}$

= Not defined (parallel to y-axis)

Slope of diagonal AC = $\frac{1-1}{-1+5}$

= 0 [parallel to x-axis]

Therefore, both the diagonals AC and BD will be perpendicular.

And the quadrilateral formed by the given points will be a rhombus.

5 0

3 years ago