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bogdanovich [222]
3 years ago
5

Wat is the value of 6 in 90.16

Mathematics
2 answers:
zhuklara [117]3 years ago
8 0
Hundredths place - the first number after a decimal is the tenths place and then the second is the hundredths place.
pshichka [43]3 years ago
5 0
Ones place i think would be the awser
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Find the Area of the shape below.
kobusy [5.1K]

Answer:

42.5 Units squared

Step-by-step explanation:

Rectangle: 9.0 * 3.5 = 31.5

Square: 2 * 2 = 4

Left Triangle: 2 * 2 * .5 = 2

Right triangle: 5 * 2 * .5 = 5

31.5 + 4 + 2 + 5 = 42.5

5 0
3 years ago
Amir says the graph of<br> y= x2 + 16 has -4 as a zero. Is Amir correct?<br> Explain.
sukhopar [10]

Answer:

  • Amir is wrong

Step-by-step explanation:

The function y = x² + 16 is positive for any value of x, hence the graph has no intersection with the x-axis and therefore has no real zero's.

If no zero's, -4 also can't be its zero.

8 0
2 years ago
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What is 345 x 12? Plz tell me the answer
katovenus [111]
The answer is 4140
——————————
3 0
3 years ago
An equation of the perpendicular bisector of the line segment with end points (3,0) and (-3,0) is
Norma-Jean [14]

Answer:

x = 0

Step-by-step explanation:

Given

(x_1,y_1) = (3,0)

(x_2,y_2) = (-3,0)

Required

The equation of the perpendicular bisector.

First, calculate the midpoint of the given endpoints

(x,y) = 0.5(x_1 + x_2, y_1 + y_2)

(x,y) = 0.5(3-3, 0+ 0)

(x,y) = 0.5(0, 0)

Open bracket

(x,y) = (0.5*0, 0.5*0)

(x,y) = (0, 0)

Next, determine the slope of the given endpoints.

m = \frac{y_2 - y_1}{x_2 - x_1}

m = \frac{0 - 0}{-3- 3}

m = \frac{0}{-6}

m = 0

Next, calculate the slope of the perpendicular bisector.

When two lines are perpendicular, the relationship between them is:

m_2 = -\frac{1}{m_1}

In this case:

m = m_1 = 0

So:

m_2 = -\frac{1}{0}

m_2 = unde\ fined

Since the slope is unde\ fined, the equation is:

x = a

Where:

(x,y) = (a,b)

Recall that:

(x,y) = (0, 0)

So:

a = 0

Hence, the equation is:

x = 0

5 0
3 years ago
HELPPPP!!!!!!!!!!!!!!!!!!!!!!
Leno4ka [110]
The answer is C. 23 is a term
4 0
3 years ago
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