All categories

2 years ago

Can someone help me with these two problems and show work please !!

Mathematics

2 answers:

Semenov [28]2 years ago

8 0

I give u the answers of first and second one.

thank you

brainlist please

pshichka [43]2 years ago

6 0

Answer:

First answer is 9 or 1 . Second answer is 7 or -7.

Step-by-step explanation:

$(x-5)^{2} =16$

$x-5 = 4$ or $x - 5 =-4$

so $x = 9$ or $x = 1$

$2x^{2} = 98$

$x^{2} =49$

$x=7$ or $x = -7$

You might be interested in

Find the area of the irregular figure

lukranit [14]

Rectangle=21 times 14=294 cm^2
Triangle=Base Times Height
Height=21-12=9cm
Area of triangle=16 times 9 =144

Area of both=294+144=438cm^2

7 0

3 years ago

Read 2 more answers

4. Millie's checking account had $98.26

SpyIntel [72]

At the end of the third week she had $136.91

5 0

3 years ago

HELP ME AGAIN PLSSSSS

Greeley [361]

Answer:

Step-by-step explanation:

4^2 = 16

4×4=16 thats the answer

5 0

3 years ago

Read 2 more answers

Evaluate the Riemann sum for f(x) = 3 - 1/2 times x between 2 and 14 where the endpoints are included with six subintervals taki

Digiron [165]

Answer:

-6

Step-by-step explanation:

Given that :

we are to evaluate the Riemann sum for $f(x) = 3 - \dfrac{1}{2}x$ from 2 ≤ x ≤ 14

where the endpoints are included with six subintervals, taking the sample points to be the left endpoints.

The Riemann sum can be computed as follows:

$L_6 = \int ^{14}_{2}3- \dfrac{1}{2}x \dx = \lim_{n \to \infty} \sum \limits ^6 _{i=1} \ f (x_i -1) \Delta x$

where:

$\Delta x = \dfrac{b-a}{a}$

a = 2

b =14

n = 6

∴

$\Delta x = \dfrac{14-2}{6}$

$\Delta x = \dfrac{12}{6}$

$\Delta x =2$

Hence;

$x_0 = 2 \\ \\ x_1 = 2+2 =4\\ \\ x_2 = 2 + 2(2) \\ \\ x_i = 2 + 2i$

Here, we are using left end-points, then:

$x_i-1 = 2+ 2(i-1)$

Replacing it into Riemann equation;

$L_6 = \lim_{n \to \infty} \sum \imits ^{6}_{i=1} \begin {pmatrix}3 - \dfrac{1}{2} \begin {pmatrix} 2+2 (i-1) \end {pmatrix} \end {pmatrix}2$

$L_6 = \lim_{n \to \infty} \sum \imits ^{6}_{i=1} 6 - (2+2(i-1))$

$L_6 = \lim_{n \to \infty} \sum \imits ^{6}_{i=1} 6 - (2+2i-2)$

$L_6 = \lim_{n \to \infty} \sum \imits ^{6}_{i=1} 6 -2i$

$L_6 = \lim_{n \to \infty} \sum \imits ^{6}_{i=1} 6 - \lim_{n \to \infty} \sum \imits ^{6}_{i=1} 2i$

$L_6 = \lim_{n \to \infty} \sum \imits ^{6}_{i=1} 6 - 2 \lim_{n \to \infty} \sum \imits ^{6}_{i=1} i$

Estimating the integrals, we have :

$= 6n - 2 ( \dfrac{n(n-1)}{2})$

= 6n - n(n+1)

replacing thevalue of n = 6 (i.e the sub interval number), we have:

= 6(6) - 6(6+1)

= 36 - 36 -6

= -6

5 0

3 years ago

What is the solution to the equation 1/3(x-2)=1/5(x+54)+2

Dennis_Churaev [7]

Answer:

x=101

Step-by-step explanation:

5 0

4 years ago

Read 2 more answers