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Snowcat [4.5K]
3 years ago
6

Indica el valor del 3 en cada uno de los siguientes números 35.097 _ 753.978 _ 7.543 _ 312.501

Mathematics
1 answer:
lara31 [8.8K]3 years ago
5 0
Free points pls and thank you
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What is 6789.987 rounded to the nearest thousandths
ivann1987 [24]
The number 6789.987 is already in rounded in thousandth
8 0
3 years ago
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Find the Y INTERCEPT and SLOPE of the line <br> 5x+3y= -4<br> Y INTERCEPT= <br> SLOPE=
Veronika [31]
Rearrange the equation so the left side starts out as y =

5x + 3y = -4
3y = -5x - 4
y = (-5/3)x - (4/3)

the slope is in front of the x and the y-intercept iis the other number
slope is -5/3
y-interxept = -4/3
6 0
3 years ago
The perimeter of a triangle with two equal sides is 46 cm. If it's base were lengthened by 3 cm and each leg were shortened by 5
lana66690 [7]

Answer:

The original triangle has sides of 18 cm, 18 cm and 10 cm

Step-by-step explanation:

Let x cm be the length of the leg of the original triangle and y cm be the length of the base of the original triangle.

1. Since the perimeter of the original triangle is 46 cm, then

x+x+y=46.

2. If it's base were lengthened by 3 cm, then it becomes y+3 cm. If each leg were shortened by 5 cm, then each leg becomes x-5 cm. All three sides became equal, then

y+3=x-5.

3. Solve the system of two equations:

\left\{\begin{array}{l}2x+y=46\\x-5=y+3\end{array}\right.\Rightarrow \left\{\begin{array}{l}2(y+8)+y=46\\x=y+8\end{array}\right.,\ 2y+y=46-16,\ 3y=30,\ y=10\ cm.

Then x=10+8=18\ cm.

3 0
3 years ago
The class size increased 25% to 40 students. What was the original class size?
vladimir1956 [14]
The original class had 32 students. 25% of 32 is 8 and 32+8=40
6 0
3 years ago
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Help with num 4 please.​
kati45 [8]

Answer:

y=1/2(x-1)

Step-by-step explanation:

If x=t^2 and t>0, then t=sqrt(x).

If t=sqrt(x) or x^(1/2) and y =1-1/t, then y=1-x^(-1/2).

The x-intercept is when y=0.

So we need to solve 0=1-x^(-1/2) to find point P.

Add x^(-1/2) on both sides: x^(-1/2)=1.

Raise both sides to -2 power: x=1

So point P is (1,0).

Let's find tangent line at point (1,0).

We will need the slope so let's differentiate.

y'=0+1/2x^(-3/2)

y'=1/(2x^(3/2))

The slope at x=1 is y'=1/(2[1]^(3/2))=1/(2×1)=1/2.

Recall point-slope form is y-y1=m(x-x1).

So our line we are looking for is y-0=1/2(x-1)

Let's simplify left hand side y=1/2(x-1)

4 0
3 years ago
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