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Goryan [66]
2 years ago
14

Help me please I will give 20 points and mark u as brainliest

Mathematics
1 answer:
hoa [83]2 years ago
3 0
The altitudes are perpendicular to the sides of the triangle based on the incenter theorem
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Please help me with these Algebra Questions!!
Bess [88]

Answer:

There is nothing here!

Step-by-step explanation:

I don't see anything to solve.

: )

3 0
3 years ago
There are 127 lbs of fruits in the store: apples, oranges, and peaches. There are four times as many apples as there are peaches
Pepsi [2]
Let
x------> amount of pounds of apples
y------> amount of pounds of oranges
z-----> amount of pounds of peaches

we know that
x+y+z=127------> equation 1
x=4z-----> equation 2
y=7+z----> equation 3

substitute  equation 2 and equation 3 in equation 1
x+y+z=127----> (4z)+(7+z)+z=127---> 6z=127-7---> 6z=120----> z=20 lb
x=4z---> x=4*20----> x=80 lb
y=7+z----> y=7+20---> y=27 lb

 amount of pounds of apples----> 80 lb
amount of pounds of oranges----> 27 lb
amount of pounds of peaches---> 20 lb

4 0
3 years ago
Help me please! thanksss
Fittoniya [83]
7:45 -1 hour=6:45

The concert started at 6:45


7:45+2 hours=9:45

The concert will be over at 9:45

The time outside the theater is

5:30a.m. or 5:30p.m.

It's most likely 5:30p.m. as 5:30a.m. is really early in the morning.
8 0
3 years ago
Using subtraction what is 4 1/6-1 5/6
krek1111 [17]

Answer:

4

Step-by-step explanation:

the answer is 4 because 2.33333333333

is 4

4 0
3 years ago
A curve is given by y=(x-a)√(x-b) for x≥b, where a and b are constants, cuts the x axis at A where x=b+1. Show that the gradient
ankoles [38]

<u>Answer:</u>

A curve is given by y=(x-a)√(x-b) for x≥b. The gradient of the curve at A is 1.

<u>Solution:</u>

We need to show that the gradient of the curve at A is 1

Here given that ,

y=(x-a) \sqrt{(x-b)}  --- equation 1

Also, according to question at point A (b+1,0)

So curve at point A will, put the value of x and y

0=(b+1-a) \sqrt{(b+1-b)}

0=b+1-c --- equation 2

According to multiple rule of Differentiation,

y^{\prime}=u^{\prime} y+y^{\prime} u

so, we get

{u}^{\prime}=1

v^{\prime}=\frac{1}{2} \sqrt{(x-b)}

y^{\prime}=1 \times \sqrt{(x-b)}+(x-a) \times \frac{1}{2} \sqrt{(x-b)}

By putting value of point A and putting value of eq 2 we get

y^{\prime}=\sqrt{(b+1-b)}+(b+1-a) \times \frac{1}{2} \sqrt{(b+1-b)}

y^{\prime}=\frac{d y}{d x}=1

Hence proved that the gradient of the curve at A is 1.

7 0
3 years ago
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