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ioda
3 years ago
13

HELP (SEE IMAGE) what triangle must be a right triangle and why???

Mathematics
2 answers:
pashok25 [27]3 years ago
8 0
The answer would be D
DIA [1.3K]3 years ago
8 0

Answer:

D. \triangle BGC is right angle because $\overleftrightarrow{EG}$⊥ $\overline{CC'}$CC'.

Step-by-step explanation:

Given A figure  in which line EG is perpendicular to Segment CC'

Therfore ,\triangle BGC is a right angled triangle because in which \angle BGC=90^{\circ}

A right angled triangle is that type of triangle in which one angle is of 90 degree .Then ,we say the the triangle is a right angled triangle.

Now , we can see from the given figure  in \triangle BGC line EG is perpendicular is given . Hence, we are given

$\overleftrightarrow{EG}$\perp$\overline {CC}'$

\therefore , \angle BGC=90^{\circ}

Hence, we can say the triangle BGC is a right triangle.

Now, we check given option

A. It is false. Because no information about the triangle A'B'C' .In given statement the triangle A'B'C' is the reflect image of the triangle ABC .So we can not say \triangle A'B'C' is a right angle .

B. ADC is a right triangle because $\overline{AA'}  intersect AC at C.

It is false.Because  we can see from given the figure AC not meet in given figure.

C. \triangle BCC' is a right triangel  because B lies of the reflection .It is false.Because we can't say that BCC' is a right angle if B lies of the reflection.

D. \triangle BGC is a right triangle  because $\overleftrightarrow{EG}\perp$\overline{CC'}

It is true.  Because EG is perpendicular it means \angle BGC=90^{\circ}. Hence, we can say triangle BGC is right triangle.

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Find the maxmium area of a rectangular plot of land which can be enclosed by rope of length 60 metres​
mario62 [17]

Answer:

225 m²

Step-by-step explanation:

If W is the width of the rectangle, and L is the length, then:

60 = 2W + 2L

A = WL

Use the first equation to solve for one of the variables:

30 = W + L

L = 30 − W

Substitute into the second equation:

A = W (30 − W)

A = 30W − W²

This is a parabola, so we can find the vertex using the formula x = -b/(2a).

W = -30 / (2 × -1)

W = 15

Or, we can use calculus:

dA/dW = 30 − 2W

0 = 30 − 2W

W = 15

Solving for L:

L = 30 − W

L = 15

So the maximum area is:

A = WL

A = (15)(15)

A = 225

3 0
3 years ago
Barry read for 15 more minutes than Robert.
Greeley [361]
Hi!

C. v + 15

Barry read for 15 MORE minutes than Robert.
6 0
3 years ago
Can someone plz help me. How can you find the inequalities of 11/15 and 5/7. Next 5/9 and 7/13. Next 11/15 and 5/7. Lastly 5/9 a
LekaFEV [45]
To make this a little clearer, let's give the pairs of inequalities the same denominator:

<span>Question 1: 
</span>\frac{11}{15} ? \frac{5}{7}
First, apply the common denominator to the first fraction:
(\frac{11}{15})7 \\  \frac{11}{15} *  \frac{7}{7}  \\  \frac{11*7}{15*7}  \\  \frac{77}{105}
Do the same for the second:15( \frac{5}{7}) \\  \frac{5}{7}* \frac{15}{15} \\  \frac{5*15}{7*15}  \\  \frac{75}{105}
Nest, compare the two fractions:
\frac{77}{105} \ \textgreater \   \frac{75}{105}
Therefore:
\frac{11}{15} > \frac{5}{7}
<span>
Question Two:</span>
\frac{5}{9} ? \frac{7}{13}
Apply the common denominator to fraction one:
13( \frac{5}{9}) \\  \frac{5}{9} * \frac{13}{13}  \\  \frac{5*13}{9*13}  \\  \frac{65}{117}
Fraction two:
9(\frac{7}{13}) \\  \frac{7}{13} *  \frac{9}{9}  \\  \frac{7*9}{13*9}  \\  \frac{63}{117}
Evaluate:
\frac{65}{117} > \frac{63}{117}
Therefore:
<span>\frac{5}{9} > \frac{7}{13}
</span>
Hope this helps!
5 0
3 years ago
Change 10 inches per second to miles per second
d1i1m1o1n [39]

Answer:

I don’t even know, what was the question?

Step-by-step explanation:

5 0
3 years ago
Read 2 more answers
Directions: Graph the equation. Label the points where the line crosses the axes.
kolezko [41]

Answer:

Both the equations landed on point ( 15.167 , 8.267 )

Step-by-step explanation:

7 0
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