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3 years ago

A rectangle and a triangle with the same base and height have the same area. True or false

Mathematics

2 answers:

Tanya [424]3 years ago

8 0

Answer:

False.

Step-by-step explanation:

The triangle has 1/2 the area of the rectangle.

marin [14]3 years ago

5 0

Answer:

<h2>FALSE</h2>

Step-by-step explanation:

The formula of an area of a rectangle:

$A_{\boxed{\ }}=lw$

l - length

w - width

The fromula of an area of a triangle:

$A_\triangle=\dfrac{bh}{2}$

b - base

h - height

If l = a and w = h, then

$A_\triangle=\dfrac{lw}{2}=\dfrac{1}{2}lw=\dfrac{1}{2}A_\boxed{\ }$

<em>The area of the triangle is equal to half the area of the rectangle with the same base and height.</em>

<em>The area of the rectangle is equal to twice the area of the triangle with the same base and height.</em>

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What is the logarithmic function modeled by the following table? x f(x) 9 2 27 3 81 4

Nataly_w [17]

Answer:

Required logarithmic function is :

$f\left(x\right)=\log_3\left(x\right)$

Step-by-step explanation:

We have been fiven that table for the logarithmic function is:

x f(x)

9 2

27 3

81 4

which can be rewritten as:

x f(x)

3^2 2

3^3 3

3^4 4

Which are basically powers of 3

So we can use logarithmic function as

$\log_3\left(9\right)=\log_3\left(3^2\right)=2\log_3\left(3\right)=2(1)=2$

Hence required logarithmic function is :

$f\left(x\right)=\log_3\left(x\right)$

8 0

3 years ago

Please could you find the answers to the questions in the attachment.

Fudgin [204]

( $\frac{x1+x2}{2} , \frac{y1+y2}{2}$ )we need 3 equations
1. midpoint equation which is ( $\frac{x1+x2}{2} , \frac{y1+y2}{2}$ ) when you have 2 points

2. distance formula which is D= $\sqrt{(x2-x1)^{2}+(y2-y1)^{2}}$

3. area of trapezoid formula whhic is (b1+b2) times 1/2 times height

so

x is midpoint of B and C
B=11,10
c=19,6
x1=11
y1=10
x2=19
y2=6
midpoint=( $\frac{11+19}{2} , \frac{10+6}{2}$ )
midpoint=( $\frac{30}{2} , \frac{16}{2}$ )
midpoint= (15,8)

point x=(15,8)

y is midpoint of A and D
A=5,8
D=21,0
x1=5
y1=8
x2=21
y2=0
midpoint=( $\frac{5+21}{2} , \frac{8+0}{2}$ )
midpoint=( $\frac{26}{2} , \frac{8}{2}$ )
midpoint=(13,4)

Y=(13,4)

legnths of BC and XY
B=(11,10)
C=(19,6)
x1=11
y1=10
x2=19
y2=6
D= $\sqrt{(19-11)^{2}+(6-10)^{2}}$
D= $\sqrt{(8)^{2}+(-4)^{2}}$
D= $\sqrt{64+16}$
D= $\sqrt{80}$
D= $4 \sqrt{5}$
BC= $4 \sqrt{5}$

X=15,8
Y=(13,4)
x1=15
y1=8
x2=13
y2=4
D= $\sqrt{(13-15)^{2}+(4-8)^{2}}$
D= $\sqrt{(-2)^{2}+(-4)^{2}}$
D= $\sqrt{4+16}$
D= $\sqrt{20}$
D= $2 \sqrt{5}$
XY= $2 \sqrt{5}$

the thingummy is a trapezoid
we need to find AD and BC and XY
we already know that BC= $4 \sqrt{5}$ and XY= $2 \sqrt{5}$

AD distance
A=5,8
D=21,0
x1=5
y1=8
x2=21
y2=0
D= $\sqrt{(21-5)^{2}+(0-8)^{2}}$
D= $\sqrt{(16)^{2}+(-8)^{2}}$
D= $\sqrt{256+64}$
D= $\sqrt{320}$
D= $4 \sqrt{2}$
AD= $4 \sqrt{2}$

so we have
AD= $4 \sqrt{2}$
BC= $4 \sqrt{5}$
XY= $2 \sqrt{5}$

AD and BC are base1 and base 2
XY=height
so
(b1+b2) times 1/2 times height
( $4 \sqrt{2}+4 \sqrt{5}$ ) times 1/2 times $2 \sqrt{5}$ =
( $4 \sqrt{2}+4 \sqrt{5}$ ) times \sqrt{5} [/tex] =
$4 \sqrt{10}+4*5$ = $4 \sqrt{10}+20$ =80 $\sqrt{10}$ =252.982

X=(15,8)
Y=(13,4)
BC= $4 \sqrt{5}$
XY= $2 \sqrt{5}$
Area=80 $\sqrt{10}$ square unit or 252.982 square units

7 0

3 years ago

Can anyone help me please?

miss Akunina [59]

Answer:

5. -18x^3y^4 +54x^2y^4

6. 6x^2 -40x +50

Step-by-step explanation:

Use the distributive property. The factor outside parentheses multiplies each term inside parentheses.

$(9xy)(-2x^2y^3+6xy)=(9xy)(-2x^2y^3)+(9xy)(6xy^3)\\\\=\boxed{-18x^3y^4+54x^2y^4}$