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Oliga [24]
3 years ago
8

The longest side of an acute isosceles triangle is 8 centimeters. rounded to the nearest tenth, what is the smallest possible le

ngth of one of the two congruent sides?
A:4.0
B:4.1
C:5.6
D:5.7
Mathematics
2 answers:
Aleksandr [31]3 years ago
8 0
It is an acute isosceles triangle, so the vertex angle is smaller than 90.
if the vertex angle is 90, then it is a 45-45-90 triangle, when the longest side (the hypotenuse) is 8, each leg is 8/√2=5.65685

Because the vertex angle needs to be smaller than 90, the two equal side is larger than 5.6568, I'd go with D.
Kobotan [32]3 years ago
4 0
4.1cm
I helped you so you're welcome
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Round to the nearest ten thousand 103,075,067
vodka [1.7K]

The number in the ten thousands place is 7

Look at the number in the place value next to it. It is a 5. Note that numbers 5 and greater round up, while 4 and lower round down.

103,080,000 is your answer

hope this helps

6 0
3 years ago
Read 2 more answers
Need help with a and b parts
Masja [62]

Answer:

a)  10/27 or 0.37037

b) -8/7

Step-by-step explanation:

a)  Enter x=-5 into both equations and divide h(-5)/g(-5).  That yields 10/27 or 0.37037037037 . . .

b) g(x) cannot be 0 (can't divide by 0.  That would occur if x = -8/7, so it (-8/7) is not in the domain.

6 0
3 years ago
Please fill in the blank
Katen [24]

Answer:

<em>The independent variable is h</em>

<em>The dependent variable is P</em>

<em>Domain of h: 3<h<23</em>

<em>Range of h: 26<P<46</em>

Step-by-step explanation:

Domain and Range of Functions

A function that is explicitly defined can have restrictions on its variables that guarantee its existence. For example, a function with a variable denominator must restrict the domain (values of the dependent variable) so, the denominator is never zero. Once determined the domain of the function, the range is made of all values the function can take.

We have a triangle with sides h,13, and 10. The perimeter of the triangle, called P is the sum of its sides

P=h+13+10=23+h

h is the independent variable, whose value modify the value of P, so P is the dependent value

Since h is the third side of the triangle, some restrictions apply. First, h cannot be less than the difference between 13 and 10, because it will produce an incomplete triangle. For example, if h=2, it won't be enough to bond the other vertices of the triangle with a line. So the minimum value of h is 3.

On the other hand, h cannot exceed the sum of the other sides, because in that case, they won't be long enough to reach the last vertex of the triangle. So h must be less than 23. It forms the restriction of the domain

3<h<23

As a consequence, the range will be restricted also.

Since P=23+h, we get h=P-23. Replacing in the above domain:

3<P-23<23

Adding 23, the range is revealed:

26<P<46

7 0
3 years ago
Read 2 more answers
Write an equation of the line that passes through the point (–1, 4) with slope 2. A. y+1=−2(x−4) B. y+1=2(x−4) C. y−4=2(x+1) D.
In-s [12.5K]

Answer:

y-4=2\,(x+1)

which agrees with answer C in your list of possible answers.

Step-by-step explanation:

We can use the general point-slope form of a line of slope m and going through the point (x_0, y_0):

y-y_0=m(x-x_0)

which in our case, given the info on the slope (2) and the point (-1, 4) becomes:

y-y_0=m\,(x-x_0)\\y-4=2\,(x-(-1))\\y-4=2\,(x+1)

6 0
3 years ago
The concentration C (in mg/dl), of an antibiotic in a patient’s bloodstream per hour, t, is given by:
lilavasa [31]

We want the concentration of the antibiotic to be at a minimum of 5 mg/dL, so we want to find the time interval, where t>0, for which C(t)\ge5.


We have


\dfrac{55t}{t^2+25}\ge5\iff55t\ge5t^2+625\iff t^2-11t+25\le0


Completing the square, we get


t^2-11t+\dfrac{121}4-\dfrac{121}4+25\le0\iff\left(t-\dfrac{11}2\right)^2\le\dfrac{21}4


\implies\left|t-\dfrac{11}2\right|\le\dfrac{\sqrt{21}}2


\implies-\dfrac{\sqrt{21}}2\le t-\dfrac{11}2\le\dfrac{\sqrt{21}}2


\implies\dfrac{11-\sqrt{21}}2\le t\le\dfrac{11+\sqrt{21}}2

4 0
3 years ago
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