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k0ka [10]
2 years ago
7

Help me please? Thank you

Mathematics
1 answer:
Fantom [35]2 years ago
5 0

23

Step-by-step explanation:

angle DEF = 74 degrees.

well I would use sins but it's quite simple

Length of DE is 23, just like AB

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Drag and drop the correct surface area to match the cube. cube with edge length = 3.2 m 12.8 m²19.2 m²51.2 m²61.44 m²
Ira Lisetskai [31]
- Surface Area of a Cube = 6(L)²
- Surface Area of this cube = 6(3.2)²
                                              = 61.44 m²
3 0
3 years ago
What is the slope of the line? y + 5=2(x+1)
mel-nik [20]

Answer:

2

Step-by-step explanation:

So we have the equation:

y+5=2(x+1)

This is in the format point-slope form, where:

y-y_1=m(x-x_1)

Here, m is the slope.

In our original equation, 2 replaces m.

Therefore, our slope is 2.

5 0
3 years ago
Read 2 more answers
Can someone plz help me with Algebra 2....... :) Will give Brainliest!!
Anarel [89]
Using logarithms property of log(x)+log(y)=log(xy)
so here, you can sum the equation to;
log((x+6)*(x-6))=2
so you can simply say that;
log_{8}((x+6)( x-6))=2
and by multiplying (x+6)*(x-6)
log_{8}(x^2-36)=2
and as you know also that;
a^{b}=c is same as log _{a}c=b
so you can simply state it as;
8^2=x^2-36
64=x^2-36
64+36=x^2
x^2=100
x=10
And you can check your work by substituting with 10 instead of x in the original function.
Hope this helps!
6 0
3 years ago
The area of the triangle is<br><br> square inches.<br> fill in the blanks
Dvinal [7]

Answer:

48sq^2

Step-by-step explanation:

(12*8)/2

96/2

48

3 0
2 years ago
In the isosceles △ABC m∠ACB=120° and AD is an altitude to leg BC . What is the distance from D to base AB , if CD=4cm?
7nadin3 [17]

Correct answer is: distance from D to AB is 6cm

Solution:-

Let us assume E is the altitude drawn from D to AB.

Given that m∠ACB=120° and ABC is isosceles which means

m∠ABC=m∠BAC = \frac{180-120}{2}=30

And AC= BC

Let AC=BC=x

Then from ΔACD , cos(∠ACD) = \frac{DC}{AC} =\frac{4}{x}

Since DCB is a straight line m∠ACD+m∠ACB =180

                                              m∠ACD = 180-m∠ACB = 60

Hence cos(60)=\frac{4}{x}

          x=\frac{4}{cos60}= 8

Now let us consider ΔBDE, sin(∠DBE) = \frac{DE}{DB} =\frac{DE}{DA+AB} = \frac{DE}{4+8}

DE = 12sin(30) = 6cm

7 0
3 years ago
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