All categories

4 years ago

Deanna estimated the product of 6.45 and 10.2 below.

Mathematics

2 answers:

Alex Ar [27]4 years ago

7 0

Answer:

b i am smart

Step-by-step explanation:

Bas_tet [7]4 years ago

5 0

Answer:

Step-by-step explanation:

I took the test

You might be interested in

Find the area of a quadrilateral ABCD in which AB = 3 cm, BC = 4 cm, CD = 4 cm, DA = 5 cm and AC = 5 cm.

melamori03 [73]

Answer:

$6+2\sqrt{21}\:\mathrm{cm^2}\approx 15.17\:\mathrm{cm^2}$

Step-by-step explanation:

The quadrilateral ABCD consists of two triangles. By adding the area of the two triangles, we get the area of the entire quadrilateral.

Vertices A, B, and C form a right triangle with legs $AB=3$ , $BC=4$ , and $AC=5$ . The two legs, 3 and 4, represent the triangle's height and base, respectively.

The area of a triangle with base $b$ and height $h$ is given by $A=\frac{1}{2}bh$ . Therefore, the area of this right triangle is:

$A=\frac{1}{2}\cdot 3\cdot 4=\frac{1}{2}\cdot 12=6\:\mathrm{cm^2}$

The other triangle is a bit trickier. Triangle $\triangle ADC$ is an isosceles triangles with sides 5, 5, and 4. To find its area, we can use Heron's Formula, given by:

$A=\sqrt{s(s-a)(s-b)(s-c)}$ , where $a$ , $b$ , and $c$ are three sides of the triangle and $s$ is the semi-perimeter ( $s=\frac{a+b+c}{2}$ ).

The semi-perimeter, $s$ , is:

$s=\frac{5+5+4}{2}=\frac{14}{2}=7$

Therefore, the area of the isosceles triangle is:

$A=\sqrt{7(7-5)(7-5)(7-4)},\\A=\sqrt{7\cdot 2\cdot 2\cdot 3},\\A=\sqrt{84}, \\A=2\sqrt{21}\:\mathrm{cm^2}$

Thus, the area of the quadrilateral is:

$6\:\mathrm{cm^2}+2\sqrt{21}\:\mathrm{cm^2}=\boxed{6+2\sqrt{21}\:\mathrm{cm^2}}$

4 0

3 years ago

In a set of 3 numbers the first is a positive integer, the second is 3 more than the first and the 3rd is a square of the second

Molodets [167]

Given:

In a set of 3 numbers the first is a positive integer, the second is 3 more than the first and the 3rd is a square of the second.

To find:

The equation for the given situation if the sum of the numbers is 77.

Solution:

a. Let the first number in the set is x.

The second is 3 more than the first. So, the second number is $(x+3)$ .

The 3rd number is a square of the second. So, the third number is $(x+3)^2$ .

Therefore, the first, second and third numbers are $x,(x+3),(x+3)^2$ respectively.

b. The sum of the numbers is 77.

First number + Second number + Third number = 77

c. So, the equation in terms of x is:

$x+(x+3)+(x+3)^2=77$

Therefore, the required equation is $x+(x+3)+(x+3)^2=77$ .

d. On simplification, we get

$x+x+3+x^2+6x+9=77$ $[\because (a+b)^2=a^2+2ab+b^2]$

$x^2+(6x+x+x)+(3+9)=77$

$x^2+8x+12=77$

Subtract 77 from both sides.

$x^2+8x+12-77=77-77$

$x^2+8x-65=0$

Therefore, the simplified form of the required equation is $x^2+8x-65=0$ .

6 0

3 years ago

How can you compare seven tenths and one half without using a model

34kurt

Hi ok make a diagram that goes horzontaly and make 9 lines to get 10 squares then make another diagram that is equal to the other one and draw one lune in the middle shade in 7/10 on the first one then shade in 1/2 in the second one

7 0

3 years ago

(6x+y)(6x-7) What is the answer?

lys-0071 [83]

= 6x + 6x - 6x × 7 + 6xy - 7y

36x2 - 42x + 6xy - 7y

6 0

3 years ago

True or False?

neonofarm [45]

$FALSE!\\The\ segments\ below\ couldn't\ form\ a\ triangle.\\\\a,b,c-lengths\ sides\ of\ a\ triangle,\ then:\\\\a+b > c\\a+c > b\\b+c > c\\\\Here:a=9;\ b=4;\ c=15\\\\a+b=9+4=13 < 15!!!!$

7 0

3 years ago