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anyanavicka [17]
3 years ago
7

Please help! Will mark brainlist!

Mathematics
2 answers:
Anna007 [38]3 years ago
7 0

Answer:

C

Step-by-step explanation:

We need to find the area of BOTH the rectangle AND the triangle. So the formula to find area of a triangle is bh\frac{1}{2\\}, or bh/2

b- base

h height

And obviously to find area of rectangles and stuff like that is just l x w

So let's find area of the rectangle, AKA the whole,

5x11=55

The area of the rectangle is 55

Now for the triangle,

The base is 5, and the height is 3. 5 x 3 = 15, and 15/2 = 7.5

The area of the triangle is 7.5

Now, subtract because we don't want the area of the triangle BUT the shaded region, so 55-7.5= 47.5 We have to add a unit so its 47.5^2, so the answer is C

Hope this helps!!

-Ketifa

harina [27]3 years ago
6 0
C

I had to do this question to
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Suppose point z is halfway between point w and x a standard numberline, and point x is halfway between points z and y. Where is
Delicious77 [7]

Answer:

The point w is located at -4/3

Step-by-step explanation:

Point z is halfway between point w and x

Point x is halfway between point z and y

Point z is located at 1/3 and point y is located at 11/3

Please refer to the number line attached

Then the x is located at the mid-point of z and y

x = \frac{z+y}{2} \\\\x = \frac{\frac{1}{3}+ \frac{11}{3} }{2}

x = 2

Since z is the mid-point of w and x

z = \frac{w+ x }{2}\\\\\frac{1}{3}  = \frac{w+ 2 }{2}\\\\\frac{2}{3}  = w+ 2 \\\\w = \frac{2}{3} - 2\\\\w = -\frac{4}{3}

Therefore, the point w is located at -4/3

6 0
3 years ago
Can you please help me with this problem ​
padilas [110]

Answer:

Yes

Step-by-step explanation:

One of the ways you can prove if a parallelogram is a parallelogram is if the sides in the shape are congruent and that's exactly what this proof is showing

Hope this image helps

6 0
3 years ago
5. Artie uses 1 1/3 yards of rope to make the bottom of his hammock stronger He uses 10 inches of rope to strengthen some areas
LenKa [72]
The correct answer is C 11/8 and 48 inches
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3 years ago
Para alavancar seus lucros, o dono de uma empresa passou a investir mais no meio digital, começando pelo seu site. Com base em u
ella [17]

Answer:

85.5 reals

Step-by-step explanation:

Aqui, usando a função de análise, queremos saber a quantidade de dinheiro que deve ser investido para fornecer o número de visualizações.

A maneira como podemos resolver isso é através da substituição. Precisamos apenas substituir F (x) na equação e resolver x, o que nos dará a idéia da quantidade de dinheiro a ser investido.

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F (x) = 40x + 80

3500 = 40x + 80

40x = 3500-80 40x = 3420 x = 3420/40

x = 85,5 reais

8 0
3 years ago
If a,b,c and d are positive real numbers such that logab=8/9, logbc=-3/4, logcd=2, find the value of logd(abc)
Eva8 [605]

We can expand the logarithm of a product as a sum of logarithms:

\log_dabc=\log_da+\log_db+\log_dc

Then using the change of base formula, we can derive the relationship

\log_xy=\dfrac{\ln y}{\ln x}=\dfrac1{\frac{\ln x}{\ln y}}=\dfrac1{\log_yx}

This immediately tells us that

\log_dc=\dfrac1{\log_cd}=\dfrac12

Notice that none of a,b,c,d can be equal to 1. This is because

\log_1x=y\implies1^{\log_1x}=1^y\implies x=1

for any choice of y. This means we can safely do the following without worrying about division by 0.

\log_db=\dfrac{\ln b}{\ln d}=\dfrac{\frac{\ln b}{\ln c}}{\frac{\ln d}{\ln c}}=\dfrac{\log_cb}{\log_cd}=\dfrac1{\log_bc\log_cd}

so that

\log_db=\dfrac1{-\frac34\cdot2}=-\dfrac23

Similarly,

\log_da=\dfrac{\ln a}{\ln d}=\dfrac{\frac{\ln a}{\ln b}}{\frac{\ln d}{\ln b}}=\dfrac{\log_ba}{\log_bd}=\dfrac{\log_db}{\log_ab}

so that

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So we end up with

\log_dabc=-\dfrac34-\dfrac23+\dfrac12=-\dfrac{11}{12}

###

Another way to do this:

\log_ab=\dfrac89\implies a^{8/9}=b\implies a=b^{9/8}

\log_bc=-\dfrac34\implies b^{-3/4}=c\implies b=c^{-4/3}

\log_cd=2\implies c^2=d\implies\log_dc^2=1\implies\log_dc=\dfrac12

Then

abc=(c^{-4/3})^{9/8}c^{-4/3}c=c^{-11/6}

So we have

\log_dabc=\log_dc^{-11/6}=-\dfrac{11}6\log_dc=-\dfrac{11}6\cdot\dfrac12=-\dfrac{11}{12}

4 0
3 years ago
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