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Bezzdna [24]
1 year ago
11

Could these triangles be congruent? yes, if ab ≅ de yes, if bc = 7 yes, if ab ≅ ef no, because the hypotenuses must have differe

nt lengths
Mathematics
2 answers:
Vika [28.1K]1 year ago
7 0

Answer:

(D) no, because the hypotenuses must have different lengths

Step-by-step explanation:

Nitella [24]1 year ago
3 0

Yes, the triangles are congruent if AB ≅ DE.

Criteria for the Congruency of two Triangles

If two triangles share the same sides and angles, they are said to be congruent triangles.

Triangles are tested for congruence using the following 4 criteria:

1) Angle-side-angle(ASA): Two angles and a side of a triangle are congruent if they are equal to two angles and the corresponding side of another triangle.

2) Side-side-side(SSS): Two triangles make congruent triangles if all three sides of one triangle are equal to three sides of another.

3) Side angle side(SAS): A triangle is congruent if its two sides and any included angles are equivalent to those of another triangle's two sides and corresponding angle.

4) Hypotenuse - leg(HL): If a triangle's hypotenuse and one of its legs are equal, then the triangle's hypotenuse and leg are also equal.

Checking for Congruency in the Given Triangles

ΔDEF and ABC have a equal angles, ∠F=∠B, as well as a pair of equal sides,  DF = AC.

Thus, in order to satisfy the SAS criterion of congruent triangles, AB = DE

Learn more about Congruent Triangles here:

brainly.com/question/22062407

#SPJ4

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slava [35]
Sweater = 400
coats = 100
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in an equation it would be
4x-x=300
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5 0
3 years ago
Identify the x-intercepts of the function below f(x)=x^2+12x+24
damaskus [11]

<u>ANSWER:  </u>

x-intercepts of  \mathrm{x}^{2}+12 \mathrm{x}+24=0 \text { are }(-6+2 \sqrt{3}),(-6-2 \sqrt{3})

<u>SOLUTION:</u>

Given, f(x)=x^{2}+12 x+24 -- eqn 1

x-intercepts of the function are the points where function touches the x-axis, which means they are zeroes of the function.

Now, let us find the zeroes using quadratic formula for f(x) = 0.

X=\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}

Here, for (1) a = 1, b= 12 and c = 24

X=\frac{-(12) \pm \sqrt{(12)^{2}-4 \times 1 \times 24}}{2 \times 1}

\begin{array}{l}{X=\frac{-12 \pm \sqrt{144-96}}{2}} \\\\ {X=\frac{-12 \pm \sqrt{48}}{2}} \\\\ {X=\frac{-12 \pm \sqrt{16 \times 3}}{2}} \\\\ {X=\frac{-12 \pm 4 \sqrt{3}}{2}} \\ {X=\frac{2(-6+2 \sqrt{3})}{2}, \frac{2(-6-2 \sqrt{3})}{2}} \\\\ {X=(-6+2 \sqrt{3}),(-6-2 \sqrt{3})}\end{array}

Hence the x-intercepts of  \mathrm{x}^{2}+12 \mathrm{x}+24=0 \text { are }(-6+2 \sqrt{3}),(-6-2 \sqrt{3})

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Answer:

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Step-by-step explanation:

Well, if 6 are accepted out of 100, that's 6% or 0.06. To find how many are accepted out of 850, multiply 850 by 0.06.

4 0
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