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3 years ago

Help me with right answer plz

Mathematics

1 answer:

givi [52]3 years ago

7 0

Answer:

13.5

Step-by-step explanation:

area: 1/2*[x1(y2 - y3) + x2(y3 - y1) + x3(y1 - y2)]

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Solve by substitution

Hunter-Best [27]

Answer:

Step-by-step explanation:

y = -9 ------(1)

9x -4y = -9 -------(2)

substitute (1) into (2)

9x - 4 (-9) = -9

9x = - 45

x = -5

there you go, x = -5 and y = -9

so. (-5 , -9 )

hope this helps

please mark it brainliest

7 0

3 years ago

Anyone know the answers to 13, 14, 15, 16

kirill115 [55]

C,g,a,j my teacher graded it with me today hope this helps! :)))

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3 years ago

Which of these inequalities is graphed below? NA w 1519 he darrer 14 li AN W UM AN ww 12+ 9 6 3 wao WA ME BAL - 15 - 12-9-6-3 -

qaws [65]

Answer:(-4,3)

Step-by-step explanation:

Here, Every equation has same slope (-4) and same y-intercept (3), so we don't need to calculate those.

Now, as coordinates are in left-portion, and also touches the line, it would be: y ≥ -4x + 3

7 0

2 years ago

Mr. Donovan bought a shirt for $18.47<br> and pants for $25.95. How much<br> money did he spend?

kap26 [50]

Answer:

$34.42

Step-by-step explanation:

18.47

25.95

--------

34.42

4 0

3 years ago

Given the rectangle abcd shown below has a total area of 72. E is in the midpoint of bc and f is the midpoint of dc. What is the

scoray [572]

Refer to the attached image.

Given the rectangle ABCD of length 'l' and height 'h'.

Therefore, CD=AB = 'l' and BC = AD = 'h'

We have to determine the area of triangle AEF.

Area of triangle AEF = Area of rectangle ABCD - Area of triangle ADF - Area of triangle ECF - Area of triangle ABE

Area of triangle ADF = $\frac{1}{2}bh$

= $\frac{1}{2}(DF \times AD)$

= $\frac{1}{2}(\frac{l}{2} \times h)$

$=\frac{lh}{4}$

Area of triangle ECF = $\frac{1}{2}bh$

= $\frac{1}{2}(CF \times CE)$

= $\frac{1}{2}(\frac{l}{2} \times \frac{h}{2})$

$=\frac{lh}{8}$

Area of triangle ABE = $\frac{1}{2}bh$

= $\frac{1}{2}(AB \times BE)$

= $\frac{1}{2}(l \times \frac{h}{2})$

$=\frac{lh}{4}$

Now, area of triangle AEF =

Area of rectangle ABCD - Area of triangle ADF - Area of triangle ECF - Area of triangle ABE

= $72 - (\frac{lh}{4} + \frac{lh}{8} + \frac{lh}{4})$

= $72 - (\frac{2lh+lh+2lh}{8})$

= $72 - (\frac{5lh}{8})$

= $72 - (\frac{5 \times 72}{8})$

$=\frac{72 \times 8 - (5 \times 72)}{8}$

= 27 units

Therefore, the area of triangle AEF is 27 units.

8 0

3 years ago