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2 years ago

5

Which angle in the isosceles triangles is the biggest?

2 answers:

IgorC [24]2 years ago

7 0

Answer:

Since the two base angles are congruent (same measure), they are each 70°. If we are given a base angle of say 45°, we know the base angles are congruent (same measure) and the interior angles of any triangle always add to 180°.

nikitadnepr [17]2 years ago

5 0

Answer:

degrees

Step-by-step explanation:

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3. Simplify the expression -3(4x -2) -2x<br><br> A. -8x<br> B.-16x<br> C.-14x -2<br> D.-14x+ 6

beks73 [17]

I got C. -14x - 2
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2 years ago

Alexus [3.1K]

<h3 /><h3> $\huge \rm༆ Answer ༄$ </h3>

To find slope we use the formula ~

$\sf \dfrac{y2 - y1}{x2 - x1}$

[ Ratio of the difference of y - coordinates of the points and the difference of x - coordinates of the points ]

let's find the slope of line passing through the given points ~

<h3>1. (10 , -9) and (-4 , -14) </h3>

$\sf \dfrac{ - 9 - ( - 14)}{10 - ( - 4)}$

$\sf \dfrac{ - 9 + 14}{10 + 4}$

$\sf \dfrac{5}{14}$

<h3>2. (16 , 6) and (-4 , -14)</h3>

$\sf \dfrac{6 - ( - 14)}{16 - ( - 4)}$

$\sf \dfrac{6 + 14}{16 + 4}$

$\sf \dfrac{20}{20}$

$\sf1$

<h3>3. (9 , -4) and (5 , -20)</h3>

$\sf \dfrac{ - 4 - ( - 20)}{9 - 5}$

$\sf \dfrac{ - 4 + 20}{4}$

$\sf \dfrac{16}{4}$

$\sf4$

<h3>4. (-13 , 20) and (-18 , 20)</h3>

$\sf \dfrac{20 - 20}{ - 13 - ( - 18)}$

$\sf\dfrac{0}{ - 13 + 18}$

$0$

I hope it helps ~

___________________________________

$꧁ \: \: \large \rm{kaul} \: ꧂$

3 0

2 years ago

Negative 12 plus 44<br> negative 19 plus 61

antiseptic1488 [7]

Your answer would be 74 if you are looking for the answer.

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3 years ago

Read 2 more answers

Suppose f(x) = x square -8x +20 find x such that f(x) =24

ch4aika [34]

X=4+-2sqaurw root of 5

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3 years ago

John looked at his bank account and noticed that he had $2,029.90 in his account. All that he could remember is that he created

Over [174]

We will have that for the previous year, he had in the account:

$m_{y6}=2029.90-\frac{2029.90\cdot6}{100}\Rightarrow m_{y6}=1908.106$

For the year previous to that one, he had:

$m_{y5}=1908.106-\frac{1908.106\cdot6}{100}\Rightarrow m_{y5}=1793.61964$

For the previous year to that, he had:

$m_{y4}=1793.61964-\frac{1793.61964\cdot6}{100}\Rightarrow m_{y4}=1686.002462$

For the year previous to that one, he had:

$m_{y3}=1686.002462-\frac{1686.002462\cdot6}{100}\Rightarrow m_{y3}=1584.842314$

For the second year that the money was in the account, there was:

$m_{y2}=1584.842314-\frac{1584.842314\cdot6}{100}\Rightarrow m_{y2}=1489.751775$

And the orinial ammount of money that was in the account was:

$m_{y1}=1489.751775-\frac{1489.751775\cdot6}{100}\Rightarrow m_{y1}=1400.366669$

From this, we know that Jhon had originally $1400.37 in the bank account.

6 0

7 months ago