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2 years ago

Claire says that she can draw an obtuse equilateral triangle. Is she correct? Explain.

2 answers:

7 0

No, because an equilateral triangle is also equiangular so all sides must be 60° (60+60+60=180) and an obtuse angle is more than 90° so the triangle would not be equilateral.

Hope I Helped (:

Ivahew [28]2 years ago

4 0

No, she is not correct. an obtuse triangle is a triangle with angles equal to more than 90. An equilateral triangle has all triangle measures 60. since 60<90, Clarie is making a false statement

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Points H and F lie on circle c What is the length of line segment GH?

Nina [5.8K]

Answer:

6 units

Step-by-step explanation:

Given: Points H and F lie on circle with center C. EG = 12, EC = 9 and ∠GEC = 90°.

To find: Length of GH.

Sol: EC = CH = 9 (Radius of the same circle are equal)

Now, GC = GH + CH

GC = GH + 9

Now In ΔEGC, using pythagoras theorem,

$(Hypotenuse)^{2} = (Base)^{2} +(Altitude)^{2}$ ......(ΔEGC is a right triangle)

$(GC)^{2} = (GE)^{2} +(EC)^{2}$

$(GH + 9)^{2} = (9)^{2} +(12)^{2}$

$(GH )^{2} + (9)^{2} + 18GH = 81 + 144$

$(GH )^{2} + 81 + 18GH = 81 + 144$

$(GH )^{2} + 18GH = 144$

Now, Let GH = <em>x</em>

$x^{2} +18x = 144$

On rearranging,

$x^{2} +18 x - 144 = 0$

$x^{2} - 6x +24x + 144 = 0$

$x (x-6) + 24 (x - 6) =0$

$(x - 6) (x + 24) = 0$

So x = 6 and x = - 24

∵ x cannot be - 24 as it will not satisfy the property of right triangle.

Therefore, the length of line segment GH = 6 units. so, Option (D) is the correct answer.

3 0

1 year ago

Read 2 more answers

The true average diameter of ball bearings of a certain type is supposed to be 0.5 in. A one-sample t test will be carried out t

Naily [24]

Answer:

Option C - Do not reject the null hypothesis. There is not sufficient evidence that the true diameter differs from 0.5 in

Step-by-step explanation:

We are given;

n = 15

t-value = 1.66

Significance level;α = 0.05

So, DF = n - 1 = 15 - 1 = 14

From the one-sample t - table attached, we can see that the p - value of 0.06 at a t-value of 1.66 and a DF of 14

Now, since the P-value is 0.06,it is greater than the significance level of 0.05. Thus we do not reject the null hypothesis. We conclude that there is not sufficient evidence that the true diameter differs from 0.5 in.