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3 years ago

14

HELP ME PLESE ,3 I NEED THIS

1 answer:

Trava [24]3 years ago

5 0

Answer:

270°

Step-by-step explanation:

$m\widehat{BC} = 360\degree - (146\degree + 90\degree) \\m\widehat{BC} = 360\degree -236\degree \\m\widehat{BC} = 124\degree \\\\m\widehat{ABC} =m\widehat {AB} + m\widehat {BC} \\m\widehat{ABC} =146\degree + 124\degree \\\huge \red {\boxed {m\widehat{ABC} =270\degree}} \\$

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Which relation is not a function?

kobusy [5.1K]

Answer:

B

Step-by-step explanation:

X cant be repeated twice or it will be a non function,

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3 years ago

Help plz and it’s not 61.6 because that’s what someone has said

bekas [8.4K]

Answer: 250 total kids and 140 girls

Step-by-step explanation:

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2 years ago

A triangular lot has sides of 215m, 185m, and 125m. Find the measures of the angles at its corners

Vlada [557]

Answer:

The measures of the angles at its corners are $59.1\°,35.4\°,85.5\°$

Step-by-step explanation:

see the attached figure to better understand the problem

step 1

Find the measure of angle A

Applying the law of cosines

$185^{2}= 215^{2}+125^{2}-2(215)(125)cos(A)$

$2(215)(125)cos(A)= 215^{2}+125^{2}-185^{2}$

$cos(A)= [215^{2}+125^{2}-185^{2}]/(2(215)(125))$ $cos(A)=0.513953$

$A=arccos(0.513953)=59.1\°$

step 2

Find the measure of angle B

Applying the law of cosines

$125^{2}= 215^{2}+185^{2}-2(215)(185)cos(B)$

$2(215)(185)cos(B)= 215^{2}+185^{2}-125^{2}$

$cos(B)= [215^{2}+185^{2}-125^{2}]/(2(215)(185))$ $cos(B)=0.81489$

$B=arccos(0.81489)=35.4\°$

step 3

Find the measure of angle C

Applying the law of cosines

$215^{2}= 125^{2}+185^{2}-2(125)(185)cos(C)$

$2(125)(185)cos(C)= 125^{2}+185^{2}-215^{2}$

$cos(C)= [125^{2}+185^{2}-215^{2}]/(2(125)(185))$ $cos(C)=0.0784$

$C=arccos(0.0784)=85.5\°$

3 0

3 years ago

Factor completely

ZanzabumX [31]

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so you get .. (x+6 )(x+4) A

4 0

3 years ago

Solving Quadratic Equations using the Square Root Property

Afina-wow [57]

Answer:

$x = -2$ or $x = -5$

Step-by-step explanation:

You need to complete the square before you can take the square root of both sides.

$x^2 + 7x + 10 = 0$

Subtract 10 from both sides.

$x^2 + 7x = -10$

To complete the square, you need to add the square of half of the x-term coefficient to both sides.

The x-term coefficient is 7. Half of that is 7/2. Square it to get 49/4. Now we add 49/4 to both sides of the equation.

$x^2 + 7x + \dfrac{49}{4} = -10 + \dfrac{49}{4}$

$(x + \dfrac{7}{2})^2 = -\dfrac{40}{4} + \dfrac{49}{4}$

$(x + \dfrac{7}{2})^2 = \dfrac{9}{4}$

Now we use the square root property, if

$x^2 = k$ , then

$x = \pm \sqrt{k}$

$x + \dfrac{7}{2} = \pm \sqrt{\dfrac{9}{4}}$

$x + \dfrac{7}{2} = \pm \dfrac{3}{2}$

$x + \dfrac{7}{2} = \dfrac{3}{2}$ or $x + \dfrac{7}{2} = -\dfrac{3}{2}$

$x = -\dfrac{4}{2}$ or $x = -\dfrac{10}{2}$

$x = -2$ or $x = -5$

5 0

3 years ago