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2 years ago

What additional information is needed to prove the triangles are congruent by the SAS Prostulate

Mathematics

1 answer:

VladimirAG [237]2 years ago

8 0

It is that the angle has to be formed by two pairs of congruent/corresponding sides of the triangle

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Simplify a - b - [C- (d - e) - f] - g).

dalvyx [7]

Answer:

a-b-C+d-e+f-g

Step-by-step explanation:

Steps

$a-b-\left(C-\left(d-e\right)-f\right)-g$

Show Steps

$-\left(d-e\right):\quad-d+e$

$=a-b-\left(C+e-d-f\right)-g$

Show Steps

$-\left(C-d+e-f\right):\quad-C+d-e+f$

$=a-b-C+d-e+f-g$

7 0

2 years ago

A weight lifter can bench press 189 kg. How many milligrams (mg) is this?

Inessa [10]

A weight lifter can bench press 189 kg. How many milligrams (mg) is this?

Answer: We know that:

$1kg=1,000,000 mg$

We are given, that a weight lifter can bench press 189 kg

Therefore, a weight lifter can bench press:

$189 \times 1,000,000 mg$

$189,000,000mg$

$18.9 \times 10^{7} mg$

Therefore, $18.9 \times 10^{7} mg$ is equal to the $189 kg$

3 0

3 years ago

Write the equation.Please help

Mamont248 [21]

Y = 3x is the answer
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8 0

2 years ago

What are the amplitude. Phase shift and midline if f(x) = 2 sin(x+pi)-4

kherson [118]

Find the missing angles using either the triangle sum theorem (a2+b2=c2) or inverse trigonometric ratios

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3 years ago

Mr. Jones' other friend Ms. O'Neill has "x" meters of fencing, what is the maximum area in square meters she can have?

Alexus [3.1K]

Answer:

$Max\ Area = \frac{x^2}{16}$

Step-by-step explanation:

Given

$P = x$ ---- the perimeter of fencing

Required

The maximum area

Let

$L \to Length$

$W \to Width$

So, we have:

$P = 2(L + W)$

This gives:

$2(L + W) = x$

Divide by 2

$L + W = \frac{x}{2}$

Make L the subject

$L = \frac{x}{2} - W$

The area (A) of the fence is:

$A = L * W$

Substitute $L = \frac{x}{2} - W$

$A = (\frac{x}{2} - W) * W$

Open bracket

$A = \frac{x}{2}W - W^2$

Differentiate with respect to W

$A' = \frac{x}{2} - 2W$

Set to 0

$\frac{x}{2} - 2W = 0$

Solve for 2W

$2W = \frac{x}{2}$

Solve for W

$W = \frac{x}{4}$

Recall that:

$L = \frac{x}{2} - W$

$L = \frac{x}{2} - \frac{x}{4}$

$L = \frac{2x- x}{4}$

$L = \frac{x}{4}$

So, the maximum area is:

$A = L * W$

$A = \frac{x}{4}*\frac{x}{4}$

$Max\ Area = \frac{x^2}{16}$

8 0

2 years ago