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2 years ago

Help me please 6 grade

Mathematics

2 answers:

Rom4ik [11]2 years ago

8 0

Answer:
Top Right
Step-by-step explanation:
easy multiplication. All multiplication really is repeted addition. these words
'3 times as many¨ just means you multiply by 3. 9 divided by 3 equals 3.
all you have to do for these types of problems that you can see the quanity these problems become really simple. in this case, count the carrots that chef A chopped and divide by 3.

Finger [1]2 years ago

3 0

Answer:

The second box ix the answer

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Joyce is trying to solve the equation y = x2 − 8x + 7 using the quadratic formula. She has made an error in one of the steps bel

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3 years ago

Read 2 more answers

A. Identify the error. *

san4es73 [151]

Answer:

Step-by-step explanation:

The range, in a layman language, is a set of values for which f(x) exist. Hence suppose our pre-image is 0-6,and since f(x)=20x+4 ,then

When x=0, f(x)= 4

When x=1, f(x)=24

When x=2, f(x)=44

When x=3, f(x)= 64

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3 years ago

Determine and prove what shape is formed for the given coordinates for ABCD, and then find the

ella [17]

Answer:

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Step-by-step explanation:

Plot points A (10,-4), B (6,-7), C (3,-3), D (7,0) on the coordinate plane.

Find slopes of the sides:

$AB:\ \dfrac{-7-(-4)}{6-10}=\dfrac{3}{4}\\ \\BC:\ \dfrac{-3-(-7)}{3-6}=-\dfrac{4}{3}\\ \\CD:\ \dfrac{0-(-3)}{7-3}=\dfrac{3}{4}\\ \\DA:\ \dfrac{-4-0}{10-7}=-\dfrac{4}{3}$

The slopes of opposite sides are the same, so opposite sides are parallel. The slopes of adjacent sides have product of -1, then adjacent sides are perpendicular.

Find the lengths of all sides:

$AB=\sqrt{(10-6)^2+(-4-(-7))^2}=\sqrt{4^2+3^2}=\sqrt{16+9}=\sqrt{25}=5\ units\\ \\BC=\sqrt{(6-3)^2+(-7-(-3))^2}=\sqrt{3^2+4^2}=\sqrt{16+9}=\sqrt{25}=5\ units\\ \\CD=\sqrt{(3-7)^2+(-3-0)^2}=\sqrt{4^2+3^2}=\sqrt{16+9}=\sqrt{25}=5\ units\\ \\AD=\sqrt{(10-7)^2+(-4-0)^2}=\sqrt{3^2+4^2}=\sqrt{16+9}=\sqrt{25}=5\ units\\ \\$

All four sides are of the same length.

Quadrilateral ABCD is a square (all sides of equal length and perpendicular)

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